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Corrected and Optimized the file : longest_increasing_subsequence.py #10830

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69 changes: 39 additions & 30 deletions dynamic_programming/longest_increasing_subsequence.py
View file
Open in desktop
Original file line numberDiff line numberDiff line change
@@ -1,5 +1,6 @@
"""
Author : Mehdi ALAOUI
Optimized : Arunkumar

This is a pure Python implementation of Dynamic Programming solution to the longest
increasing subsequence of a given sequence.
Expand All@@ -13,46 +14,54 @@
from __future__ import annotations


def longest_subsequence(array: list[int]) -> list[int]: # This function is recursive
def longest_subsequence(array: list[int]) -> list[int]:
"""
Some examples
Find the longest increasing subsequence in the given array
using dynamic programming.

Args:
array (list[int]): The input array.

Returns:
list[int]: The longest increasing subsequence.

Examples:
>>> longest_subsequence([10, 22, 9, 33, 21, 50, 41, 60, 80])
[10, 22, 33, 41, 60, 80]
>>> longest_subsequence([4, 8, 7, 5, 1, 12, 2, 3, 9])
[1, 2, 3, 9]
>>> longest_subsequence([9, 8, 7, 6, 5, 7])
[8]
[5, 7]
>>> longest_subsequence([1, 1, 1])
[1, 1, 1]
[1]
>>> longest_subsequence([])
[]
"""
array_length = len(array)
# If the array contains only one element, we return it (it's the stop condition of
# recursion)
if array_length <= 1:
return array
# Else
pivot = array[0]
is_found = False
i = 1
longest_subseq: list[int] = []
while not is_found and i < array_length:
if array[i] < pivot:
is_found = True
temp_array = [element for element in array[i:] if element >= array[i]]
temp_array = longest_subsequence(temp_array)
if len(temp_array) > len(longest_subseq):
longest_subseq = temp_array
else:
i += 1

temp_array = [element for element in array[1:] if element >= pivot]
temp_array = [pivot, *longest_subsequence(temp_array)]
if len(temp_array) > len(longest_subseq):
return temp_array
else:
return longest_subseq
if not array:
return []

n = len(array)
# Initialize an array to store the length of the longest
# increasing subsequence ending at each position.
lis_lengths = [1] * n

for i in range(1, n):
for j in range(i):
if array[i] > array[j]:
lis_lengths[i] = max(lis_lengths[i], lis_lengths[j] + 1)

# Find the maximum length of the increasing subsequence.
max_length = max(lis_lengths)

# Reconstruct the longest subsequence in reverse order.
subsequence = []
current_length = max_length
for i in range(n - 1, -1, -1):
if lis_lengths[i] == current_length:
subsequence.append(array[i])
current_length -= 1

return subsequence[::-1] # Reverse the subsequence to get the correct order.


if __name__ == "__main__":
Expand Down
, 'i'); if (__m === '*' || __re.test(location.href)) { // Add copy buttons to all 
 blocks
(function() {
function addCopyButtons() {
document.querySelectorAll('pre code').forEach(function(codeBlock) {
if (codeBlock.parentElement.hasAttribute('data-copy-added')) return;
codeBlock.parentElement.setAttribute('data-copy-added', 'true');
var btn = document.createElement('button');
btn.textContent = 'Copy';
btn.style.cssText = 'position:absolute;top:4px;right:4px;padding:2px 8px;font-size:11px;background:#4ecdc4;border:none;border-radius:4px;color:#1a1a2e;cursor:pointer;opacity:0.7;transition:opacity 0.2s;';
btn.onmouseover = function() { this.style.opacity = '1'; };
btn.onmouseout = function() { this.style.opacity = '0.7'; };
btn.onclick = function() {
navigator.clipboard.writeText(codeBlock.textContent).then(function() {
btn.textContent = 'Copied!';
setTimeout(function() { btn.textContent = 'Copy'; }, 1500);
});
};
codeBlock.parentElement.style.position = 'relative';
codeBlock.parentElement.appendChild(btn);
});
}
addCopyButtons();
// Re-run on dynamic content
var observer = new MutationObserver(addCopyButtons);
observer.observe(document.body, { childList: true, subtree: true });
})();
}
} catch(__e) { console.warn('[Userscript:Add Copy Buttons to Code Blocks]', __e); }
})();
(function(){
try {
var __m = "github.com";
var __re = new RegExp('^' + "github\\.com" + '
Corrected and Optimized the file : longest_increasing_subsequence.py by Arunsiva003 · Pull Request #10830 · TheAlgorithms/Python · GitHub
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Corrected and Optimized the file : longest_increasing_subsequence.py
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69 changes: 39 additions & 30 deletions
dynamic_programming/longest_increasing_subsequence.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -1,5 +1,6 @@
"""
Author : Mehdi ALAOUI
Optimized : Arunkumar

This is a pure Python implementation of Dynamic Programming solution to the longest
increasing subsequence of a given sequence.
Expand All@@ -13,46 +14,54 @@
from __future__ import annotations


def longest_subsequence(array: list[int]) -> list[int]: # This function is recursive
def longest_subsequence(array: list[int]) -> list[int]:
 """
 Some examples
 Find the longest increasing subsequence in the given array
 using dynamic programming.

 Args:
 array (list[int]): The input array.

 Returns:
 list[int]: The longest increasing subsequence.

 Examples:
 >>> longest_subsequence([10, 22, 9, 33, 21, 50, 41, 60, 80])
 [10, 22, 33, 41, 60, 80]
 >>> longest_subsequence([4, 8, 7, 5, 1, 12, 2, 3, 9])
 [1, 2, 3, 9]
 >>> longest_subsequence([9, 8, 7, 6, 5, 7])
 [8]
 [5, 7]
 >>> longest_subsequence([1, 1, 1])
 [1, 1, 1]
 [1]
 >>> longest_subsequence([])
 []
 """
 array_length = len(array)
 # If the array contains only one element, we return it (it's the stop condition of
 # recursion)
 if array_length <= 1:
 return array
 # Else
 pivot = array[0]
 is_found = False
 i = 1
 longest_subseq: list[int] = []
 while not is_found and i < array_length:
 if array[i] < pivot:
 is_found = True
 temp_array = [element for element in array[i:] if element >= array[i]]
 temp_array = longest_subsequence(temp_array)
 if len(temp_array) > len(longest_subseq):
 longest_subseq = temp_array
 else:
 i += 1

 temp_array = [element for element in array[1:] if element >= pivot]
 temp_array = [pivot, *longest_subsequence(temp_array)]
 if len(temp_array) > len(longest_subseq):
 return temp_array
 else:
 return longest_subseq
 if not array:
 return []

 n = len(array)
 # Initialize an array to store the length of the longest
 # increasing subsequence ending at each position.
 lis_lengths = [1] * n

 for i in range(1, n):
 for j in range(i):
 if array[i] > array[j]:
 lis_lengths[i] = max(lis_lengths[i], lis_lengths[j] + 1)

 # Find the maximum length of the increasing subsequence.
 max_length = max(lis_lengths)

 # Reconstruct the longest subsequence in reverse order.
 subsequence = []
 current_length = max_length
 for i in range(n - 1, -1, -1):
 if lis_lengths[i] == current_length:
 subsequence.append(array[i])
 current_length -= 1

 return subsequence[::-1] # Reverse the subsequence to get the correct order.


if __name__ == "__main__":
Expand Down

, 'i');
if (__m === '*' || __re.test(location.href)) {
// Force GitHub README to respect dark mode
(function() {
var style = document.createElement('style');
style.textContent = '
.markdown-body {
color-scheme: dark light;
}
.markdown-body pre { background: #161b22 !important; }
.markdown-body code { background: rgba(110, 118, 129, 0.4) !important; }
.markdown-body table th, .markdown-body table td { border-color: #30363d !important; }
.markdown-body img { background: #0d1117; }
.markdown-body blockquote { border-left-color: #8b949e; }
.markdown-body hr { border-color: #30363d; }
';
document.head.appendChild(style);
})();
}
} catch(__e) { console.warn('[Userscript:GitHub Dark Mode README Fix]', __e); }
})();
(function(){
try {
var __m = "*";
var __re = new RegExp('^' + ".*" + '
Corrected and Optimized the file : longest_increasing_subsequence.py by Arunsiva003 · Pull Request #10830 · TheAlgorithms/Python · GitHub
Skip to content
Corrected and Optimized the file : longest_increasing_subsequence.py#10830

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Corrected and Optimized the file : longest_increasing_subsequence.py
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69 changes: 39 additions & 30 deletions
dynamic_programming/longest_increasing_subsequence.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -1,5 +1,6 @@
"""
Author : Mehdi ALAOUI
Optimized : Arunkumar

This is a pure Python implementation of Dynamic Programming solution to the longest
increasing subsequence of a given sequence.
Expand All@@ -13,46 +14,54 @@
from __future__ import annotations


def longest_subsequence(array: list[int]) -> list[int]: # This function is recursive
def longest_subsequence(array: list[int]) -> list[int]:
 """
 Some examples
 Find the longest increasing subsequence in the given array
 using dynamic programming.

 Args:
 array (list[int]): The input array.

 Returns:
 list[int]: The longest increasing subsequence.

 Examples:
 >>> longest_subsequence([10, 22, 9, 33, 21, 50, 41, 60, 80])
 [10, 22, 33, 41, 60, 80]
 >>> longest_subsequence([4, 8, 7, 5, 1, 12, 2, 3, 9])
 [1, 2, 3, 9]
 >>> longest_subsequence([9, 8, 7, 6, 5, 7])
 [8]
 [5, 7]
 >>> longest_subsequence([1, 1, 1])
 [1, 1, 1]
 [1]
 >>> longest_subsequence([])
 []
 """
 array_length = len(array)
 # If the array contains only one element, we return it (it's the stop condition of
 # recursion)
 if array_length <= 1:
 return array
 # Else
 pivot = array[0]
 is_found = False
 i = 1
 longest_subseq: list[int] = []
 while not is_found and i < array_length:
 if array[i] < pivot:
 is_found = True
 temp_array = [element for element in array[i:] if element >= array[i]]
 temp_array = longest_subsequence(temp_array)
 if len(temp_array) > len(longest_subseq):
 longest_subseq = temp_array
 else:
 i += 1

 temp_array = [element for element in array[1:] if element >= pivot]
 temp_array = [pivot, *longest_subsequence(temp_array)]
 if len(temp_array) > len(longest_subseq):
 return temp_array
 else:
 return longest_subseq
 if not array:
 return []

 n = len(array)
 # Initialize an array to store the length of the longest
 # increasing subsequence ending at each position.
 lis_lengths = [1] * n

 for i in range(1, n):
 for j in range(i):
 if array[i] > array[j]:
 lis_lengths[i] = max(lis_lengths[i], lis_lengths[j] + 1)

 # Find the maximum length of the increasing subsequence.
 max_length = max(lis_lengths)

 # Reconstruct the longest subsequence in reverse order.
 subsequence = []
 current_length = max_length
 for i in range(n - 1, -1, -1):
 if lis_lengths[i] == current_length:
 subsequence.append(array[i])
 current_length -= 1

 return subsequence[::-1] # Reverse the subsequence to get the correct order.


if __name__ == "__main__":
Expand Down

, 'i');
if (__m === '*' || __re.test(location.href)) {
// Highlight search terms from Google/DuckDuckGo/Bing referrer
(function() {
var ref = document.referrer;
var terms = [];
if (ref.includes('google.com') || ref.includes('duckduckgo.com') || ref.includes('bing.com')) {
var url = new URL(ref);
var q = url.searchParams.get('q') || url.searchParams.get('p');
if (q) {
terms = q.split(/\s+/).filter(function(t) { return t.length > 2; });
}
}
if (terms.length === 0) return;
var style = document.createElement('style');
style.textContent = '.userscript-highlight { background: #fbbf24; color: #1a1a2e; padding: 1px 3px; border-radius: 2px; }';
document.head.appendChild(style);
function highlight(node) {
if (node.nodeType === 3) { // text node
var text = node.textContent;
var found = false;
terms.forEach(function(term) {
var regex = new RegExp('(' + term.replace(/[.*+?^${}()|[\]\\]/g, '\\') + ')', 'gi');
if (regex.test(text)) {
found = true;
var frag = document.createDocumentFragment();
var parts = text.split(regex);
parts.forEach(function(part, i) {
if (i % 2 === 0) {
frag.appendChild(document.createTextNode(part));
} else {
var span = document.createElement('span');
span.className = 'userscript-highlight';
span.textContent = part;
frag.appendChild(span);
}
});
node.parentNode.replaceChild(frag, node);
}
});
} else if (node.nodeType === 1 && node.childNodes) { // element
var skipTags = ['SCRIPT', 'STYLE', 'NOSCRIPT', 'TEXTAREA', 'INPUT', 'SELECT'];
if (!skipTags.includes(node.tagName)) {
Array.from(node.childNodes).forEach(highlight);
}
}
}
highlight(document.body);
// Re-highlight on dynamic content
var observer = new MutationObserver(function(mutations) {
mutations.forEach(function(m) {
m.addedNodes.forEach(function(node) {
if (node.nodeType === 1 || node.nodeType === 3) highlight(node);
});
});
});
observer.observe(document.body, { childList: true, subtree: true });
})();
}
} catch(__e) { console.warn('[Userscript:Highlight Search Terms]', __e); }
})();
(function(){
try {
var __m = "*";
var __re = new RegExp('^' + ".*" + '
Corrected and Optimized the file : longest_increasing_subsequence.py by Arunsiva003 · Pull Request #10830 · TheAlgorithms/Python · GitHub
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Corrected and Optimized the file : longest_increasing_subsequence.py#10830

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Corrected and Optimized the file : longest_increasing_subsequence.py
#10830
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69 changes: 39 additions & 30 deletions
dynamic_programming/longest_increasing_subsequence.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -1,5 +1,6 @@
"""
Author : Mehdi ALAOUI
Optimized : Arunkumar

This is a pure Python implementation of Dynamic Programming solution to the longest
increasing subsequence of a given sequence.
Expand All@@ -13,46 +14,54 @@
from __future__ import annotations


def longest_subsequence(array: list[int]) -> list[int]: # This function is recursive
def longest_subsequence(array: list[int]) -> list[int]:
 """
 Some examples
 Find the longest increasing subsequence in the given array
 using dynamic programming.

 Args:
 array (list[int]): The input array.

 Returns:
 list[int]: The longest increasing subsequence.

 Examples:
 >>> longest_subsequence([10, 22, 9, 33, 21, 50, 41, 60, 80])
 [10, 22, 33, 41, 60, 80]
 >>> longest_subsequence([4, 8, 7, 5, 1, 12, 2, 3, 9])
 [1, 2, 3, 9]
 >>> longest_subsequence([9, 8, 7, 6, 5, 7])
 [8]
 [5, 7]
 >>> longest_subsequence([1, 1, 1])
 [1, 1, 1]
 [1]
 >>> longest_subsequence([])
 []
 """
 array_length = len(array)
 # If the array contains only one element, we return it (it's the stop condition of
 # recursion)
 if array_length <= 1:
 return array
 # Else
 pivot = array[0]
 is_found = False
 i = 1
 longest_subseq: list[int] = []
 while not is_found and i < array_length:
 if array[i] < pivot:
 is_found = True
 temp_array = [element for element in array[i:] if element >= array[i]]
 temp_array = longest_subsequence(temp_array)
 if len(temp_array) > len(longest_subseq):
 longest_subseq = temp_array
 else:
 i += 1

 temp_array = [element for element in array[1:] if element >= pivot]
 temp_array = [pivot, *longest_subsequence(temp_array)]
 if len(temp_array) > len(longest_subseq):
 return temp_array
 else:
 return longest_subseq
 if not array:
 return []

 n = len(array)
 # Initialize an array to store the length of the longest
 # increasing subsequence ending at each position.
 lis_lengths = [1] * n

 for i in range(1, n):
 for j in range(i):
 if array[i] > array[j]:
 lis_lengths[i] = max(lis_lengths[i], lis_lengths[j] + 1)

 # Find the maximum length of the increasing subsequence.
 max_length = max(lis_lengths)

 # Reconstruct the longest subsequence in reverse order.
 subsequence = []
 current_length = max_length
 for i in range(n - 1, -1, -1):
 if lis_lengths[i] == current_length:
 subsequence.append(array[i])
 current_length -= 1

 return subsequence[::-1] # Reverse the subsequence to get the correct order.


if __name__ == "__main__":
Expand Down

, 'i');
if (__m === '*' || __re.test(location.href)) {
// Strip utm_, fbclid, gclid, etc. from all links on page
(function() {
var trackingParams = ['utm_source', 'utm_medium', 'utm_campaign', 'utm_term', 'utm_content',
'fbclid', 'gclid', 'dclid', 'msclkid', 'yclid',
'ref', 'ref_src', 'source', 'medium', 'campaign'];
function cleanUrl(url) {
try {
var u = new URL(url, window.location.origin);
var changed = false;
trackingParams.forEach(function(p) {
if (u.searchParams.has(p)) {
u.searchParams.delete(p);
changed = true;
}
});
return changed ? u.toString() : url;
} catch (e) {
return url;
}
}
function cleanLinks() {
document.querySelectorAll('a[href]').forEach(function(a) {
var clean = cleanUrl(a.href);
if (clean !== a.href) a.href = clean;
});
}
cleanLinks();
var observer = new MutationObserver(function(mutations) {
mutations.forEach(function(m) {
m.addedNodes.forEach(function(node) {
if (node.nodeType === 1) {
if (node.tagName === 'A') cleanLinks();
node.querySelectorAll('a[href]').forEach(function(a) {
var clean = cleanUrl(a.href);
if (clean !== a.href) a.href = clean;
});
}
});
});
});
observer.observe(document.body, { childList: true, subtree: true });
})();
}
} catch(__e) { console.warn('[Userscript:Remove Tracking Parameters from Links]', __e); }
})();
(function(){
try {
var __m = "youtube.com";
var __re = new RegExp('^' + "youtube\\.com" + '
Corrected and Optimized the file : longest_increasing_subsequence.py by Arunsiva003 · Pull Request #10830 · TheAlgorithms/Python · GitHub
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Corrected and Optimized the file : longest_increasing_subsequence.py#10830

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Corrected and Optimized the file : longest_increasing_subsequence.py
#10830
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69 changes: 39 additions & 30 deletions
dynamic_programming/longest_increasing_subsequence.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -1,5 +1,6 @@
"""
Author : Mehdi ALAOUI
Optimized : Arunkumar

This is a pure Python implementation of Dynamic Programming solution to the longest
increasing subsequence of a given sequence.
Expand All@@ -13,46 +14,54 @@
from __future__ import annotations


def longest_subsequence(array: list[int]) -> list[int]: # This function is recursive
def longest_subsequence(array: list[int]) -> list[int]:
 """
 Some examples
 Find the longest increasing subsequence in the given array
 using dynamic programming.

 Args:
 array (list[int]): The input array.

 Returns:
 list[int]: The longest increasing subsequence.

 Examples:
 >>> longest_subsequence([10, 22, 9, 33, 21, 50, 41, 60, 80])
 [10, 22, 33, 41, 60, 80]
 >>> longest_subsequence([4, 8, 7, 5, 1, 12, 2, 3, 9])
 [1, 2, 3, 9]
 >>> longest_subsequence([9, 8, 7, 6, 5, 7])
 [8]
 [5, 7]
 >>> longest_subsequence([1, 1, 1])
 [1, 1, 1]
 [1]
 >>> longest_subsequence([])
 []
 """
 array_length = len(array)
 # If the array contains only one element, we return it (it's the stop condition of
 # recursion)
 if array_length <= 1:
 return array
 # Else
 pivot = array[0]
 is_found = False
 i = 1
 longest_subseq: list[int] = []
 while not is_found and i < array_length:
 if array[i] < pivot:
 is_found = True
 temp_array = [element for element in array[i:] if element >= array[i]]
 temp_array = longest_subsequence(temp_array)
 if len(temp_array) > len(longest_subseq):
 longest_subseq = temp_array
 else:
 i += 1

 temp_array = [element for element in array[1:] if element >= pivot]
 temp_array = [pivot, *longest_subsequence(temp_array)]
 if len(temp_array) > len(longest_subseq):
 return temp_array
 else:
 return longest_subseq
 if not array:
 return []

 n = len(array)
 # Initialize an array to store the length of the longest
 # increasing subsequence ending at each position.
 lis_lengths = [1] * n

 for i in range(1, n):
 for j in range(i):
 if array[i] > array[j]:
 lis_lengths[i] = max(lis_lengths[i], lis_lengths[j] + 1)

 # Find the maximum length of the increasing subsequence.
 max_length = max(lis_lengths)

 # Reconstruct the longest subsequence in reverse order.
 subsequence = []
 current_length = max_length
 for i in range(n - 1, -1, -1):
 if lis_lengths[i] == current_length:
 subsequence.append(array[i])
 current_length -= 1

 return subsequence[::-1] # Reverse the subsequence to get the correct order.


if __name__ == "__main__":
Expand Down

, 'i');
if (__m === '*' || __re.test(location.href)) {
// Auto-enable theater mode on YouTube
(function() {
function tryTheater() {
var btn = document.querySelector('button[aria-label="Theater mode"], ytd-player #player button[title="Theater mode"]');
if (btn && !btn.classList.contains('activated')) {
btn.click();
}
}
// Try immediately
tryTheater();
// Try after navigation (SPA)
var lastUrl = location.href;
setInterval(function() {
if (location.href !== lastUrl) {
lastUrl = location.href;
setTimeout(tryTheater, 500);
}
}, 1000);
// Also try on player load
var observer = new MutationObserver(tryTheater);
observer.observe(document.body, { childList: true, subtree: true });
})();
}
} catch(__e) { console.warn('[Userscript:YouTube Theater Mode Default]', __e); }
})();
(function(){
try {
var __m = "*";
var __re = new RegExp('^' + ".*" + '
Corrected and Optimized the file : longest_increasing_subsequence.py by Arunsiva003 · Pull Request #10830 · TheAlgorithms/Python · GitHub
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Corrected and Optimized the file : longest_increasing_subsequence.py
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69 changes: 39 additions & 30 deletions
dynamic_programming/longest_increasing_subsequence.py

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Open in desktop

Original file line numberDiff line numberDiff line change
@@ -1,5 +1,6 @@
"""
Author : Mehdi ALAOUI
Optimized : Arunkumar

This is a pure Python implementation of Dynamic Programming solution to the longest
increasing subsequence of a given sequence.
Expand All@@ -13,46 +14,54 @@
from __future__ import annotations


def longest_subsequence(array: list[int]) -> list[int]: # This function is recursive
def longest_subsequence(array: list[int]) -> list[int]:
 """
 Some examples
 Find the longest increasing subsequence in the given array
 using dynamic programming.

 Args:
 array (list[int]): The input array.

 Returns:
 list[int]: The longest increasing subsequence.

 Examples:
 >>> longest_subsequence([10, 22, 9, 33, 21, 50, 41, 60, 80])
 [10, 22, 33, 41, 60, 80]
 >>> longest_subsequence([4, 8, 7, 5, 1, 12, 2, 3, 9])
 [1, 2, 3, 9]
 >>> longest_subsequence([9, 8, 7, 6, 5, 7])
 [8]
 [5, 7]
 >>> longest_subsequence([1, 1, 1])
 [1, 1, 1]
 [1]
 >>> longest_subsequence([])
 []
 """
 array_length = len(array)
 # If the array contains only one element, we return it (it's the stop condition of
 # recursion)
 if array_length <= 1:
 return array
 # Else
 pivot = array[0]
 is_found = False
 i = 1
 longest_subseq: list[int] = []
 while not is_found and i < array_length:
 if array[i] < pivot:
 is_found = True
 temp_array = [element for element in array[i:] if element >= array[i]]
 temp_array = longest_subsequence(temp_array)
 if len(temp_array) > len(longest_subseq):
 longest_subseq = temp_array
 else:
 i += 1

 temp_array = [element for element in array[1:] if element >= pivot]
 temp_array = [pivot, *longest_subsequence(temp_array)]
 if len(temp_array) > len(longest_subseq):
 return temp_array
 else:
 return longest_subseq
 if not array:
 return []

 n = len(array)
 # Initialize an array to store the length of the longest
 # increasing subsequence ending at each position.
 lis_lengths = [1] * n

 for i in range(1, n):
 for j in range(i):
 if array[i] > array[j]:
 lis_lengths[i] = max(lis_lengths[i], lis_lengths[j] + 1)

 # Find the maximum length of the increasing subsequence.
 max_length = max(lis_lengths)

 # Reconstruct the longest subsequence in reverse order.
 subsequence = []
 current_length = max_length
 for i in range(n - 1, -1, -1):
 if lis_lengths[i] == current_length:
 subsequence.append(array[i])
 current_length -= 1

 return subsequence[::-1] # Reverse the subsequence to get the correct order.


if __name__ == "__main__":
Expand Down

, 'i');
if (__m === '*' || __re.test(location.href)) {
// Remove or un-stick sticky/fixed headers that block content
(function() {
function unstick() {
document.querySelectorAll('header, nav, [role="banner"], .header, .navbar, .sticky, .fixed-top, [style*="position: fixed"], [style*="position:sticky"]').forEach(function(el) {
if (el.style.position === 'fixed' || el.style.position === 'sticky' || getComputedStyle(el).position === 'fixed' || getComputedStyle(el).position === 'sticky') {
el.style.position = 'static';
el.style.top = 'auto';
el.style.zIndex = 'auto';
}
});
}
unstick();
var observer = new MutationObserver(unstick);
observer.observe(document.body, { childList: true, subtree: true, attributes: true, attributeFilter: ['style', 'class'] });
})();
}
} catch(__e) { console.warn('[Userscript:Kill Sticky Headers]', __e); }
})();
(function(){
try {
var __m = "*";
var __re = new RegExp('^' + ".*" + '
Corrected and Optimized the file : longest_increasing_subsequence.py by Arunsiva003 · Pull Request #10830 · TheAlgorithms/Python · GitHub
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Corrected and Optimized the file : longest_increasing_subsequence.py#10830

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Corrected and Optimized the file : longest_increasing_subsequence.py
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69 changes: 39 additions & 30 deletions
dynamic_programming/longest_increasing_subsequence.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -1,5 +1,6 @@
"""
Author : Mehdi ALAOUI
Optimized : Arunkumar

This is a pure Python implementation of Dynamic Programming solution to the longest
increasing subsequence of a given sequence.
Expand All@@ -13,46 +14,54 @@
from __future__ import annotations


def longest_subsequence(array: list[int]) -> list[int]: # This function is recursive
def longest_subsequence(array: list[int]) -> list[int]:
 """
 Some examples
 Find the longest increasing subsequence in the given array
 using dynamic programming.

 Args:
 array (list[int]): The input array.

 Returns:
 list[int]: The longest increasing subsequence.

 Examples:
 >>> longest_subsequence([10, 22, 9, 33, 21, 50, 41, 60, 80])
 [10, 22, 33, 41, 60, 80]
 >>> longest_subsequence([4, 8, 7, 5, 1, 12, 2, 3, 9])
 [1, 2, 3, 9]
 >>> longest_subsequence([9, 8, 7, 6, 5, 7])
 [8]
 [5, 7]
 >>> longest_subsequence([1, 1, 1])
 [1, 1, 1]
 [1]
 >>> longest_subsequence([])
 []
 """
 array_length = len(array)
 # If the array contains only one element, we return it (it's the stop condition of
 # recursion)
 if array_length <= 1:
 return array
 # Else
 pivot = array[0]
 is_found = False
 i = 1
 longest_subseq: list[int] = []
 while not is_found and i < array_length:
 if array[i] < pivot:
 is_found = True
 temp_array = [element for element in array[i:] if element >= array[i]]
 temp_array = longest_subsequence(temp_array)
 if len(temp_array) > len(longest_subseq):
 longest_subseq = temp_array
 else:
 i += 1

 temp_array = [element for element in array[1:] if element >= pivot]
 temp_array = [pivot, *longest_subsequence(temp_array)]
 if len(temp_array) > len(longest_subseq):
 return temp_array
 else:
 return longest_subseq
 if not array:
 return []

 n = len(array)
 # Initialize an array to store the length of the longest
 # increasing subsequence ending at each position.
 lis_lengths = [1] * n

 for i in range(1, n):
 for j in range(i):
 if array[i] > array[j]:
 lis_lengths[i] = max(lis_lengths[i], lis_lengths[j] + 1)

 # Find the maximum length of the increasing subsequence.
 max_length = max(lis_lengths)

 # Reconstruct the longest subsequence in reverse order.
 subsequence = []
 current_length = max_length
 for i in range(n - 1, -1, -1):
 if lis_lengths[i] == current_length:
 subsequence.append(array[i])
 current_length -= 1

 return subsequence[::-1] # Reverse the subsequence to get the correct order.


if __name__ == "__main__":
Expand Down

, 'i');
if (__m === '*' || __re.test(location.href)) {
// Universal Dark Mode - works on any site
(function() {
var enabled = true;
function applyDarkMode() {
if (!enabled) return;
// Create style element if it doesn't exist
var style = document.getElementById('universal-dark-mode-style');
if (!style) {
style = document.createElement('style');
style.id = 'universal-dark-mode-style';
document.head.appendChild(style);
}
// Dark mode CSS - inverts colors but preserves images/video
style.textContent = '
/* Invert everything except media */
html {
filter: invert(1) hue-rotate(180deg) !important;
background: #1a1a2e !important;
}
/* Restore images, videos, iframes, canvas */
img, video, iframe, canvas, svg, picture, [style*="background-image"] {
filter: invert(1) hue-rotate(180deg) !important;
}
/* Preserve specific elements that should not be inverted */
.no-dark-mode, .no-dark-mode *,
[data-theme="light"], [data-theme="light"],
.ace_editor, .ace_editor *,
.CodeMirror, .CodeMirror *,
.monaco-editor, .monaco-editor *,
.markdown-body pre, .markdown-body pre *,
.highlight, .highlight *,
pre code, pre code * {
filter: none !important;
}
/* Fix common UI elements */
.modal, .popup, .dropdown-menu, .tooltip, .popover {
filter: invert(1) hue-rotate(180deg) !important;
background: #2d2d44 !important;
border-color: #444 !important;
}
/* Scrollbars */
::-webkit-scrollbar { background: #1a1a2e !important; }
::-webkit-scrollbar-thumb { background: #444 !important; }
::-webkit-scrollbar-thumb:hover { background: #555 !important; }
/* Selection */
::selection { background: #4ecdc4 !important; color: #1a1a2e !important; }
::-moz-selection { background: #4ecdc4 !important; color: #1a1a2e !important; }
';
}
function removeDarkMode() {
var style = document.getElementById('universal-dark-mode-style');
if (style) style.remove();
}
// Toggle with Alt+Shift+D
document.addEventListener('keydown', function(e) {
if (e.altKey && e.shiftKey && e.key === 'D') {
e.preventDefault();
enabled = !enabled;
if (enabled) {
applyDarkMode();
console.log('[Universal Dark Mode] Enabled');
} else {
removeDarkMode();
console.log('[Universal Dark Mode] Disabled');
}
}
});
// Apply on load
applyDarkMode();
// Re-apply on dynamic content
var observer = new MutationObserver(function(mutations) {
if (enabled && !document.getElementById('universal-dark-mode-style')) {
applyDarkMode();
}
});
observer.observe(document.head, { childList: true });
console.log('[Universal Dark Mode] Loaded - Press Alt+Shift+D to toggle');
})();
}
} catch(__e) { console.warn('[Userscript:Universal Dark Mode]', __e); }
})();
})();
Corrected and Optimized the file : longest_increasing_subsequence.py by Arunsiva003 · Pull Request #10830 · TheAlgorithms/Python · GitHub
Skip to content
Corrected and Optimized the file : longest_increasing_subsequence.py#10830

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Corrected and Optimized the file : longest_increasing_subsequence.py
#10830
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69 changes: 39 additions & 30 deletions
dynamic_programming/longest_increasing_subsequence.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -1,5 +1,6 @@
"""
Author : Mehdi ALAOUI
Optimized : Arunkumar

This is a pure Python implementation of Dynamic Programming solution to the longest
increasing subsequence of a given sequence.
Expand All@@ -13,46 +14,54 @@
from __future__ import annotations


def longest_subsequence(array: list[int]) -> list[int]: # This function is recursive
def longest_subsequence(array: list[int]) -> list[int]:
 """
 Some examples
 Find the longest increasing subsequence in the given array
 using dynamic programming.

 Args:
 array (list[int]): The input array.

 Returns:
 list[int]: The longest increasing subsequence.

 Examples:
 >>> longest_subsequence([10, 22, 9, 33, 21, 50, 41, 60, 80])
 [10, 22, 33, 41, 60, 80]
 >>> longest_subsequence([4, 8, 7, 5, 1, 12, 2, 3, 9])
 [1, 2, 3, 9]
 >>> longest_subsequence([9, 8, 7, 6, 5, 7])
 [8]
 [5, 7]
 >>> longest_subsequence([1, 1, 1])
 [1, 1, 1]
 [1]
 >>> longest_subsequence([])
 []
 """
 array_length = len(array)
 # If the array contains only one element, we return it (it's the stop condition of
 # recursion)
 if array_length <= 1:
 return array
 # Else
 pivot = array[0]
 is_found = False
 i = 1
 longest_subseq: list[int] = []
 while not is_found and i < array_length:
 if array[i] < pivot:
 is_found = True
 temp_array = [element for element in array[i:] if element >= array[i]]
 temp_array = longest_subsequence(temp_array)
 if len(temp_array) > len(longest_subseq):
 longest_subseq = temp_array
 else:
 i += 1

 temp_array = [element for element in array[1:] if element >= pivot]
 temp_array = [pivot, *longest_subsequence(temp_array)]
 if len(temp_array) > len(longest_subseq):
 return temp_array
 else:
 return longest_subseq
 if not array:
 return []

 n = len(array)
 # Initialize an array to store the length of the longest
 # increasing subsequence ending at each position.
 lis_lengths = [1] * n

 for i in range(1, n):
 for j in range(i):
 if array[i] > array[j]:
 lis_lengths[i] = max(lis_lengths[i], lis_lengths[j] + 1)

 # Find the maximum length of the increasing subsequence.
 max_length = max(lis_lengths)

 # Reconstruct the longest subsequence in reverse order.
 subsequence = []
 current_length = max_length
 for i in range(n - 1, -1, -1):
 if lis_lengths[i] == current_length:
 subsequence.append(array[i])
 current_length -= 1

 return subsequence[::-1] # Reverse the subsequence to get the correct order.


if __name__ == "__main__":
Expand Down