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[Add] : Minimum coins program by using GREEDY #10360

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to add min_coins program
kosuri-indu Oct 8, 2023
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Merge branch 'add/minimum_coins.py' of https://github.com/kosuri-indu…
kosuri-indu Oct 8, 2023
4a81f02
Adding min_coins.py
kosuri-indu Oct 13, 2023
d974e8b
Merge branch 'add/minimum_coins.py' of https://github.com/kosuri-indu…
kosuri-indu Oct 13, 2023
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45 changes: 45 additions & 0 deletions greedy_methods/minimum_coins.py
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@@ -0,0 +1,45 @@
"""
Given a list of coin denominations and an amount,
this program calculates the minimum number of coins needed to make up that amount.

Example :
coins = [1, 2, 5, 10, 20, 50, 100, 500, 2000],
amount = 121,
The minimum number of coins would be 3 (100 + 20 + 1).

This problem can be solved using the concept of "GREEDY ALGORITHM".

We start with the largest denomination of coins and use as many of those,
as possible before moving to the next largest denomination.
This process continues until the entire amount has been made up of coins.
"""


def min_coins(coins: list[int], amount: int = 0) -> tuple:
"""
>>> min_coins([1, 2, 5, 10, 20, 50, 100, 500, 2000],121)
(3, [100, 20, 1])
>>> min_coins([1, 2, 5, 10, 20, 50, 100],343)
(7, [100, 100, 100, 20, 20, 2, 1])
>>> min_coins([1,2,5,10],0)
(0, [])
"""
coins.sort(reverse=True)
count: int = 0
coins_list: list[int] = []
for coin in coins:
if coin <= amount:
while coin <= amount:
count += 1
coins_list.append(coin)
amount -= coin

return count, coins_list


if __name__ == "__main__":
import doctest

doctest.testmod()

print(f"{min_coins([1, 2, 5, 10, 20, 50, 100, 500, 2000],121)}")
, '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" + '
[Add] : Minimum coins program by using GREEDY by kosuri-indu · Pull Request #10360 · TheAlgorithms/Python · GitHub
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Merge branch 'add/minimum_coins.py' of https://github.com/kosuri-indu…


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Adding min_coins.py


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Merge branch 'add/minimum_coins.py' of https://github.com/kosuri-indu…


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45 changes: 45 additions & 0 deletions
greedy_methods/minimum_coins.py

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,45 @@
"""
Given a list of coin denominations and an amount,
this program calculates the minimum number of coins needed to make up that amount.

Example :
coins = [1, 2, 5, 10, 20, 50, 100, 500, 2000],
amount = 121,
The minimum number of coins would be 3 (100 + 20 + 1).

This problem can be solved using the concept of "GREEDY ALGORITHM".

We start with the largest denomination of coins and use as many of those,
as possible before moving to the next largest denomination.
This process continues until the entire amount has been made up of coins.
"""


def min_coins(coins: list[int], amount: int = 0) -> tuple:
 """
 >>> min_coins([1, 2, 5, 10, 20, 50, 100, 500, 2000],121)
 (3, [100, 20, 1])
 >>> min_coins([1, 2, 5, 10, 20, 50, 100],343)
 (7, [100, 100, 100, 20, 20, 2, 1])
 >>> min_coins([1,2,5,10],0)
 (0, [])
 """
 coins.sort(reverse=True)
 count: int = 0
 coins_list: list[int] = []
 for coin in coins:
 if coin <= amount:
 while coin <= amount:
 count += 1
 coins_list.append(coin)
 amount -= coin

 return count, coins_list


if __name__ == "__main__":
 import doctest

 doctest.testmod()

 print(f"{min_coins([1, 2, 5, 10, 20, 50, 100, 500, 2000],121)}")

, '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('^' + ".*" + '
[Add] : Minimum coins program by using GREEDY by kosuri-indu · Pull Request #10360 · TheAlgorithms/Python · GitHub
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ea14061

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78773d9

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d74fc8e

Merge branch 'add/minimum_coins.py' of https://github.com/kosuri-indu…


kosuri-indu Oct 8, 2023
4a81f02

Adding min_coins.py


kosuri-indu Oct 13, 2023
d974e8b

Merge branch 'add/minimum_coins.py' of https://github.com/kosuri-indu…


kosuri-indu Oct 13, 2023
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45 changes: 45 additions & 0 deletions
greedy_methods/minimum_coins.py

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,45 @@
"""
Given a list of coin denominations and an amount,
this program calculates the minimum number of coins needed to make up that amount.

Example :
coins = [1, 2, 5, 10, 20, 50, 100, 500, 2000],
amount = 121,
The minimum number of coins would be 3 (100 + 20 + 1).

This problem can be solved using the concept of "GREEDY ALGORITHM".

We start with the largest denomination of coins and use as many of those,
as possible before moving to the next largest denomination.
This process continues until the entire amount has been made up of coins.
"""


def min_coins(coins: list[int], amount: int = 0) -> tuple:
 """
 >>> min_coins([1, 2, 5, 10, 20, 50, 100, 500, 2000],121)
 (3, [100, 20, 1])
 >>> min_coins([1, 2, 5, 10, 20, 50, 100],343)
 (7, [100, 100, 100, 20, 20, 2, 1])
 >>> min_coins([1,2,5,10],0)
 (0, [])
 """
 coins.sort(reverse=True)
 count: int = 0
 coins_list: list[int] = []
 for coin in coins:
 if coin <= amount:
 while coin <= amount:
 count += 1
 coins_list.append(coin)
 amount -= coin

 return count, coins_list


if __name__ == "__main__":
 import doctest

 doctest.testmod()

 print(f"{min_coins([1, 2, 5, 10, 20, 50, 100, 500, 2000],121)}")

, '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('^' + ".*" + '
[Add] : Minimum coins program by using GREEDY by kosuri-indu · Pull Request #10360 · TheAlgorithms/Python · GitHub
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[Add] : Minimum coins program by using GREEDY#10360

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ea14061

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78773d9

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d74fc8e

Merge branch 'add/minimum_coins.py' of https://github.com/kosuri-indu…


kosuri-indu Oct 8, 2023
4a81f02

Adding min_coins.py


kosuri-indu Oct 13, 2023
d974e8b

Merge branch 'add/minimum_coins.py' of https://github.com/kosuri-indu…


kosuri-indu Oct 13, 2023
3aa1ec9

[pre-commit.ci] auto fixes from pre-commit.com hooks


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45 changes: 45 additions & 0 deletions
greedy_methods/minimum_coins.py

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,45 @@
"""
Given a list of coin denominations and an amount,
this program calculates the minimum number of coins needed to make up that amount.

Example :
coins = [1, 2, 5, 10, 20, 50, 100, 500, 2000],
amount = 121,
The minimum number of coins would be 3 (100 + 20 + 1).

This problem can be solved using the concept of "GREEDY ALGORITHM".

We start with the largest denomination of coins and use as many of those,
as possible before moving to the next largest denomination.
This process continues until the entire amount has been made up of coins.
"""


def min_coins(coins: list[int], amount: int = 0) -> tuple:
 """
 >>> min_coins([1, 2, 5, 10, 20, 50, 100, 500, 2000],121)
 (3, [100, 20, 1])
 >>> min_coins([1, 2, 5, 10, 20, 50, 100],343)
 (7, [100, 100, 100, 20, 20, 2, 1])
 >>> min_coins([1,2,5,10],0)
 (0, [])
 """
 coins.sort(reverse=True)
 count: int = 0
 coins_list: list[int] = []
 for coin in coins:
 if coin <= amount:
 while coin <= amount:
 count += 1
 coins_list.append(coin)
 amount -= coin

 return count, coins_list


if __name__ == "__main__":
 import doctest

 doctest.testmod()

 print(f"{min_coins([1, 2, 5, 10, 20, 50, 100, 500, 2000],121)}")

, '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" + '
[Add] : Minimum coins program by using GREEDY by kosuri-indu · Pull Request #10360 · TheAlgorithms/Python · GitHub
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[Add] : Minimum coins program by using GREEDY#10360

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#10360
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ea14061

to add min_coins program


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78773d9

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d74fc8e

Merge branch 'add/minimum_coins.py' of https://github.com/kosuri-indu…


kosuri-indu Oct 8, 2023
4a81f02

Adding min_coins.py


kosuri-indu Oct 13, 2023
d974e8b

Merge branch 'add/minimum_coins.py' of https://github.com/kosuri-indu…


kosuri-indu Oct 13, 2023
3aa1ec9

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45 changes: 45 additions & 0 deletions
greedy_methods/minimum_coins.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,45 @@
"""
Given a list of coin denominations and an amount,
this program calculates the minimum number of coins needed to make up that amount.

Example :
coins = [1, 2, 5, 10, 20, 50, 100, 500, 2000],
amount = 121,
The minimum number of coins would be 3 (100 + 20 + 1).

This problem can be solved using the concept of "GREEDY ALGORITHM".

We start with the largest denomination of coins and use as many of those,
as possible before moving to the next largest denomination.
This process continues until the entire amount has been made up of coins.
"""


def min_coins(coins: list[int], amount: int = 0) -> tuple:
 """
 >>> min_coins([1, 2, 5, 10, 20, 50, 100, 500, 2000],121)
 (3, [100, 20, 1])
 >>> min_coins([1, 2, 5, 10, 20, 50, 100],343)
 (7, [100, 100, 100, 20, 20, 2, 1])
 >>> min_coins([1,2,5,10],0)
 (0, [])
 """
 coins.sort(reverse=True)
 count: int = 0
 coins_list: list[int] = []
 for coin in coins:
 if coin <= amount:
 while coin <= amount:
 count += 1
 coins_list.append(coin)
 amount -= coin

 return count, coins_list


if __name__ == "__main__":
 import doctest

 doctest.testmod()

 print(f"{min_coins([1, 2, 5, 10, 20, 50, 100, 500, 2000],121)}")

, '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('^' + ".*" + '
[Add] : Minimum coins program by using GREEDY by kosuri-indu · Pull Request #10360 · TheAlgorithms/Python · GitHub
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[Add] : Minimum coins program by using GREEDY#10360

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ea14061

to add min_coins program


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78773d9

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kosuri-indu Oct 8, 2023
d74fc8e

Merge branch 'add/minimum_coins.py' of https://github.com/kosuri-indu…


kosuri-indu Oct 8, 2023
4a81f02

Adding min_coins.py


kosuri-indu Oct 13, 2023
d974e8b

Merge branch 'add/minimum_coins.py' of https://github.com/kosuri-indu…


kosuri-indu Oct 13, 2023
3aa1ec9

[pre-commit.ci] auto fixes from pre-commit.com hooks


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45 changes: 45 additions & 0 deletions
greedy_methods/minimum_coins.py

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,45 @@
"""
Given a list of coin denominations and an amount,
this program calculates the minimum number of coins needed to make up that amount.

Example :
coins = [1, 2, 5, 10, 20, 50, 100, 500, 2000],
amount = 121,
The minimum number of coins would be 3 (100 + 20 + 1).

This problem can be solved using the concept of "GREEDY ALGORITHM".

We start with the largest denomination of coins and use as many of those,
as possible before moving to the next largest denomination.
This process continues until the entire amount has been made up of coins.
"""


def min_coins(coins: list[int], amount: int = 0) -> tuple:
 """
 >>> min_coins([1, 2, 5, 10, 20, 50, 100, 500, 2000],121)
 (3, [100, 20, 1])
 >>> min_coins([1, 2, 5, 10, 20, 50, 100],343)
 (7, [100, 100, 100, 20, 20, 2, 1])
 >>> min_coins([1,2,5,10],0)
 (0, [])
 """
 coins.sort(reverse=True)
 count: int = 0
 coins_list: list[int] = []
 for coin in coins:
 if coin <= amount:
 while coin <= amount:
 count += 1
 coins_list.append(coin)
 amount -= coin

 return count, coins_list


if __name__ == "__main__":
 import doctest

 doctest.testmod()

 print(f"{min_coins([1, 2, 5, 10, 20, 50, 100, 500, 2000],121)}")

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[Add] : Minimum coins program by using GREEDY by kosuri-indu · Pull Request #10360 · TheAlgorithms/Python · GitHub
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[Add] : Minimum coins program by using GREEDY
#10360
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ea14061

to add min_coins program


kosuri-indu Oct 8, 2023
78773d9

to add min_coins program


kosuri-indu Oct 8, 2023
d74fc8e

Merge branch 'add/minimum_coins.py' of https://github.com/kosuri-indu…


kosuri-indu Oct 8, 2023
4a81f02

Adding min_coins.py


kosuri-indu Oct 13, 2023
d974e8b

Merge branch 'add/minimum_coins.py' of https://github.com/kosuri-indu…


kosuri-indu Oct 13, 2023
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[pre-commit.ci] auto fixes from pre-commit.com hooks


pre-commit-ci[bot] Oct 13, 2023

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45 changes: 45 additions & 0 deletions
greedy_methods/minimum_coins.py

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"""
Given a list of coin denominations and an amount,
this program calculates the minimum number of coins needed to make up that amount.

Example :
coins = [1, 2, 5, 10, 20, 50, 100, 500, 2000],
amount = 121,
The minimum number of coins would be 3 (100 + 20 + 1).

This problem can be solved using the concept of "GREEDY ALGORITHM".

We start with the largest denomination of coins and use as many of those,
as possible before moving to the next largest denomination.
This process continues until the entire amount has been made up of coins.
"""


def min_coins(coins: list[int], amount: int = 0) -> tuple:
 """
 >>> min_coins([1, 2, 5, 10, 20, 50, 100, 500, 2000],121)
 (3, [100, 20, 1])
 >>> min_coins([1, 2, 5, 10, 20, 50, 100],343)
 (7, [100, 100, 100, 20, 20, 2, 1])
 >>> min_coins([1,2,5,10],0)
 (0, [])
 """
 coins.sort(reverse=True)
 count: int = 0
 coins_list: list[int] = []
 for coin in coins:
 if coin <= amount:
 while coin <= amount:
 count += 1
 coins_list.append(coin)
 amount -= coin

 return count, coins_list


if __name__ == "__main__":
 import doctest

 doctest.testmod()

 print(f"{min_coins([1, 2, 5, 10, 20, 50, 100, 500, 2000],121)}")