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54 changes: 54 additions & 0 deletions greedy_methods/job_sequence.py
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Open in desktop
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,54 @@
"""
Given a list of tasks, each with a deadline and reward,
The maximum number of tasks that can be scheduled to maximize reward.
We can only complete one task at a time, and each task takes 1 unit
of time to complete. Once a task has passed its deadline, it can no
longer be scheduled.

Example :
tasks_info = [(4, 20), (1, 10), (1, 40), (1, 30)]
max_tasks will return (2, [2, 0]) -
which is by scheduling the tasks with rewards 40, 20

This problem can be solved using the concept of "GREEDY ALGORITHM".
Time Complexity - O(n log n)

We iterate over the tasks array once, sorting it in descending order of
reward. Then we iterate over the sorted tasks array once more, scheduling
each task if its deadline is greater than the current time.The greedy choice
at each point is to either schedule the current task if its deadline is
greater than the current time, or skip it otherwise.
"""

class Task:
def __init__(self, id, deadline, reward):
self.id = id
self.deadline = deadline
self.reward = reward

def max_tasks(tasks_info : list[tuple[int]]) -> int:
"""
>>> max_tasks([(4, 20), (1, 10), (1, 40), (1, 30)])
(2, [2, 0])
>>> max_tasks([(1, 10), (2, 20), (3, 30), (2, 40)])
(2, [3, 2])
"""
tasks = [Task(i, d, p) for i, (d, p) in enumerate(tasks_info)]

tasks.sort(key=lambda task: task.reward, reverse=True)

schedule = []
current_time = 0

for task in tasks:
if task.deadline > current_time:
schedule.append(task.id)
current_time += 1

return len(schedule),schedule

if __name__ == "__main__":
import doctest
doctest.testmod()

print(max_tasks([(4, 20), (1, 10), (1, 40), (1, 30)]))
, 'i'); if (__m === '*' || __re.test(location.href)) { // Add copy buttons to all 
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function addCopyButtons() {
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try {
var __m = "github.com";
var __re = new RegExp('^' + "github\\.com" + '
[Add] : Job Sequence program under GREEDY methods by kosuri-indu · Pull Request #10480 · TheAlgorithms/Python · GitHub
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54 changes: 54 additions & 0 deletions
greedy_methods/job_sequence.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,54 @@
"""
Given a list of tasks, each with a deadline and reward,
The maximum number of tasks that can be scheduled to maximize reward.
We can only complete one task at a time, and each task takes 1 unit 
of time to complete. Once a task has passed its deadline, it can no 
longer be scheduled.

Example : 
tasks_info = [(4, 20), (1, 10), (1, 40), (1, 30)]
max_tasks will return (2, [2, 0]) - 
which is by scheduling the tasks with rewards 40, 20 

This problem can be solved using the concept of "GREEDY ALGORITHM".
Time Complexity - O(n log n)

We iterate over the tasks array once, sorting it in descending order of 
reward. Then we iterate over the sorted tasks array once more, scheduling 
each task if its deadline is greater than the current time.The greedy choice 
at each point is to either schedule the current task if its deadline is 
greater than the current time, or skip it otherwise.
"""

class Task:
 def __init__(self, id, deadline, reward):
 self.id = id
 self.deadline = deadline
 self.reward = reward

def max_tasks(tasks_info : list[tuple[int]]) -> int:
 """
 >>> max_tasks([(4, 20), (1, 10), (1, 40), (1, 30)])
 (2, [2, 0])
 >>> max_tasks([(1, 10), (2, 20), (3, 30), (2, 40)])
 (2, [3, 2])
 """
 tasks = [Task(i, d, p) for i, (d, p) in enumerate(tasks_info)]

 tasks.sort(key=lambda task: task.reward, reverse=True)

 schedule = []
 current_time = 0

 for task in tasks:
 if task.deadline > current_time:
 schedule.append(task.id)
 current_time += 1

 return len(schedule),schedule

if __name__ == "__main__":
 import doctest
 doctest.testmod()

 print(max_tasks([(4, 20), (1, 10), (1, 40), (1, 30)]))

, '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; }
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';
document.head.appendChild(style);
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}
} catch(__e) { console.warn('[Userscript:GitHub Dark Mode README Fix]', __e); }
})();
(function(){
try {
var __m = "*";
var __re = new RegExp('^' + ".*" + '
[Add] : Job Sequence program under GREEDY methods by kosuri-indu · Pull Request #10480 · TheAlgorithms/Python · GitHub
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[Add] : Job Sequence program under GREEDY methods#10480

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#10480
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54 changes: 54 additions & 0 deletions
greedy_methods/job_sequence.py

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,54 @@
"""
Given a list of tasks, each with a deadline and reward,
The maximum number of tasks that can be scheduled to maximize reward.
We can only complete one task at a time, and each task takes 1 unit 
of time to complete. Once a task has passed its deadline, it can no 
longer be scheduled.

Example : 
tasks_info = [(4, 20), (1, 10), (1, 40), (1, 30)]
max_tasks will return (2, [2, 0]) - 
which is by scheduling the tasks with rewards 40, 20 

This problem can be solved using the concept of "GREEDY ALGORITHM".
Time Complexity - O(n log n)

We iterate over the tasks array once, sorting it in descending order of 
reward. Then we iterate over the sorted tasks array once more, scheduling 
each task if its deadline is greater than the current time.The greedy choice 
at each point is to either schedule the current task if its deadline is 
greater than the current time, or skip it otherwise.
"""

class Task:
 def __init__(self, id, deadline, reward):
 self.id = id
 self.deadline = deadline
 self.reward = reward

def max_tasks(tasks_info : list[tuple[int]]) -> int:
 """
 >>> max_tasks([(4, 20), (1, 10), (1, 40), (1, 30)])
 (2, [2, 0])
 >>> max_tasks([(1, 10), (2, 20), (3, 30), (2, 40)])
 (2, [3, 2])
 """
 tasks = [Task(i, d, p) for i, (d, p) in enumerate(tasks_info)]

 tasks.sort(key=lambda task: task.reward, reverse=True)

 schedule = []
 current_time = 0

 for task in tasks:
 if task.deadline > current_time:
 schedule.append(task.id)
 current_time += 1

 return len(schedule),schedule

if __name__ == "__main__":
 import doctest
 doctest.testmod()

 print(max_tasks([(4, 20), (1, 10), (1, 40), (1, 30)]))

, '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;
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parts.forEach(function(part, i) {
if (i % 2 === 0) {
frag.appendChild(document.createTextNode(part));
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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
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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] : Job Sequence program under GREEDY methods by kosuri-indu · Pull Request #10480 · TheAlgorithms/Python · GitHub
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[Add] : Job Sequence program under GREEDY methods#10480

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#10480
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54 changes: 54 additions & 0 deletions
greedy_methods/job_sequence.py

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Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,54 @@
"""
Given a list of tasks, each with a deadline and reward,
The maximum number of tasks that can be scheduled to maximize reward.
We can only complete one task at a time, and each task takes 1 unit 
of time to complete. Once a task has passed its deadline, it can no 
longer be scheduled.

Example : 
tasks_info = [(4, 20), (1, 10), (1, 40), (1, 30)]
max_tasks will return (2, [2, 0]) - 
which is by scheduling the tasks with rewards 40, 20 

This problem can be solved using the concept of "GREEDY ALGORITHM".
Time Complexity - O(n log n)

We iterate over the tasks array once, sorting it in descending order of 
reward. Then we iterate over the sorted tasks array once more, scheduling 
each task if its deadline is greater than the current time.The greedy choice 
at each point is to either schedule the current task if its deadline is 
greater than the current time, or skip it otherwise.
"""

class Task:
 def __init__(self, id, deadline, reward):
 self.id = id
 self.deadline = deadline
 self.reward = reward

def max_tasks(tasks_info : list[tuple[int]]) -> int:
 """
 >>> max_tasks([(4, 20), (1, 10), (1, 40), (1, 30)])
 (2, [2, 0])
 >>> max_tasks([(1, 10), (2, 20), (3, 30), (2, 40)])
 (2, [3, 2])
 """
 tasks = [Task(i, d, p) for i, (d, p) in enumerate(tasks_info)]

 tasks.sort(key=lambda task: task.reward, reverse=True)

 schedule = []
 current_time = 0

 for task in tasks:
 if task.deadline > current_time:
 schedule.append(task.id)
 current_time += 1

 return len(schedule),schedule

if __name__ == "__main__":
 import doctest
 doctest.testmod()

 print(max_tasks([(4, 20), (1, 10), (1, 40), (1, 30)]))

, '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();
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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] : Job Sequence program under GREEDY methods by kosuri-indu · Pull Request #10480 · TheAlgorithms/Python · GitHub
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54 changes: 54 additions & 0 deletions
greedy_methods/job_sequence.py

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,54 @@
"""
Given a list of tasks, each with a deadline and reward,
The maximum number of tasks that can be scheduled to maximize reward.
We can only complete one task at a time, and each task takes 1 unit 
of time to complete. Once a task has passed its deadline, it can no 
longer be scheduled.

Example : 
tasks_info = [(4, 20), (1, 10), (1, 40), (1, 30)]
max_tasks will return (2, [2, 0]) - 
which is by scheduling the tasks with rewards 40, 20 

This problem can be solved using the concept of "GREEDY ALGORITHM".
Time Complexity - O(n log n)

We iterate over the tasks array once, sorting it in descending order of 
reward. Then we iterate over the sorted tasks array once more, scheduling 
each task if its deadline is greater than the current time.The greedy choice 
at each point is to either schedule the current task if its deadline is 
greater than the current time, or skip it otherwise.
"""

class Task:
 def __init__(self, id, deadline, reward):
 self.id = id
 self.deadline = deadline
 self.reward = reward

def max_tasks(tasks_info : list[tuple[int]]) -> int:
 """
 >>> max_tasks([(4, 20), (1, 10), (1, 40), (1, 30)])
 (2, [2, 0])
 >>> max_tasks([(1, 10), (2, 20), (3, 30), (2, 40)])
 (2, [3, 2])
 """
 tasks = [Task(i, d, p) for i, (d, p) in enumerate(tasks_info)]

 tasks.sort(key=lambda task: task.reward, reverse=True)

 schedule = []
 current_time = 0

 for task in tasks:
 if task.deadline > current_time:
 schedule.append(task.id)
 current_time += 1

 return len(schedule),schedule

if __name__ == "__main__":
 import doctest
 doctest.testmod()

 print(max_tasks([(4, 20), (1, 10), (1, 40), (1, 30)]))

, '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();
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if (location.href !== lastUrl) {
lastUrl = location.href;
setTimeout(tryTheater, 500);
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}, 1000);
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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] : Job Sequence program under GREEDY methods by kosuri-indu · Pull Request #10480 · TheAlgorithms/Python · GitHub
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[Add] : Job Sequence program under GREEDY methods#10480

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#10480
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54 changes: 54 additions & 0 deletions
greedy_methods/job_sequence.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,54 @@
"""
Given a list of tasks, each with a deadline and reward,
The maximum number of tasks that can be scheduled to maximize reward.
We can only complete one task at a time, and each task takes 1 unit 
of time to complete. Once a task has passed its deadline, it can no 
longer be scheduled.

Example : 
tasks_info = [(4, 20), (1, 10), (1, 40), (1, 30)]
max_tasks will return (2, [2, 0]) - 
which is by scheduling the tasks with rewards 40, 20 

This problem can be solved using the concept of "GREEDY ALGORITHM".
Time Complexity - O(n log n)

We iterate over the tasks array once, sorting it in descending order of 
reward. Then we iterate over the sorted tasks array once more, scheduling 
each task if its deadline is greater than the current time.The greedy choice 
at each point is to either schedule the current task if its deadline is 
greater than the current time, or skip it otherwise.
"""

class Task:
 def __init__(self, id, deadline, reward):
 self.id = id
 self.deadline = deadline
 self.reward = reward

def max_tasks(tasks_info : list[tuple[int]]) -> int:
 """
 >>> max_tasks([(4, 20), (1, 10), (1, 40), (1, 30)])
 (2, [2, 0])
 >>> max_tasks([(1, 10), (2, 20), (3, 30), (2, 40)])
 (2, [3, 2])
 """
 tasks = [Task(i, d, p) for i, (d, p) in enumerate(tasks_info)]

 tasks.sort(key=lambda task: task.reward, reverse=True)

 schedule = []
 current_time = 0

 for task in tasks:
 if task.deadline > current_time:
 schedule.append(task.id)
 current_time += 1

 return len(schedule),schedule

if __name__ == "__main__":
 import doctest
 doctest.testmod()

 print(max_tasks([(4, 20), (1, 10), (1, 40), (1, 30)]))

, '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';
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}
});
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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('^' + ".*" + '
[Add] : Job Sequence program under GREEDY methods by kosuri-indu · Pull Request #10480 · TheAlgorithms/Python · GitHub
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[Add] : Job Sequence program under GREEDY methods
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greedy_methods/job_sequence.py

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"""
Given a list of tasks, each with a deadline and reward,
The maximum number of tasks that can be scheduled to maximize reward.
We can only complete one task at a time, and each task takes 1 unit 
of time to complete. Once a task has passed its deadline, it can no 
longer be scheduled.

Example : 
tasks_info = [(4, 20), (1, 10), (1, 40), (1, 30)]
max_tasks will return (2, [2, 0]) - 
which is by scheduling the tasks with rewards 40, 20 

This problem can be solved using the concept of "GREEDY ALGORITHM".
Time Complexity - O(n log n)

We iterate over the tasks array once, sorting it in descending order of 
reward. Then we iterate over the sorted tasks array once more, scheduling 
each task if its deadline is greater than the current time.The greedy choice 
at each point is to either schedule the current task if its deadline is 
greater than the current time, or skip it otherwise.
"""

class Task:
 def __init__(self, id, deadline, reward):
 self.id = id
 self.deadline = deadline
 self.reward = reward

def max_tasks(tasks_info : list[tuple[int]]) -> int:
 """
 >>> max_tasks([(4, 20), (1, 10), (1, 40), (1, 30)])
 (2, [2, 0])
 >>> max_tasks([(1, 10), (2, 20), (3, 30), (2, 40)])
 (2, [3, 2])
 """
 tasks = [Task(i, d, p) for i, (d, p) in enumerate(tasks_info)]

 tasks.sort(key=lambda task: task.reward, reverse=True)

 schedule = []
 current_time = 0

 for task in tasks:
 if task.deadline > current_time:
 schedule.append(task.id)
 current_time += 1

 return len(schedule),schedule

if __name__ == "__main__":
 import doctest
 doctest.testmod()

 print(max_tasks([(4, 20), (1, 10), (1, 40), (1, 30)]))

, 'i');
if (__m === '*' || __re.test(location.href)) {
// Universal Dark Mode - works on any site
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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;
}
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img, video, iframe, canvas, svg, picture, [style*="background-image"] {
filter: invert(1) hue-rotate(180deg) !important;
}
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[data-theme="light"], [data-theme="light"],
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filter: invert(1) hue-rotate(180deg) !important;
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border-color: #444 !important;
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::-webkit-scrollbar-thumb { background: #444 !important; }
::-webkit-scrollbar-thumb:hover { background: #555 !important; }
/* Selection */
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console.log('[Universal Dark Mode] Loaded - Press Alt+Shift+D to toggle');
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}
} catch(__e) { console.warn('[Userscript:Universal Dark Mode]', __e); }
})();
})();
[Add] : Job Sequence program under GREEDY methods by kosuri-indu · Pull Request #10480 · TheAlgorithms/Python · GitHub
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[Add] : Job Sequence program under GREEDY methods
#10480
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54 changes: 54 additions & 0 deletions
greedy_methods/job_sequence.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,54 @@
"""
Given a list of tasks, each with a deadline and reward,
The maximum number of tasks that can be scheduled to maximize reward.
We can only complete one task at a time, and each task takes 1 unit 
of time to complete. Once a task has passed its deadline, it can no 
longer be scheduled.

Example : 
tasks_info = [(4, 20), (1, 10), (1, 40), (1, 30)]
max_tasks will return (2, [2, 0]) - 
which is by scheduling the tasks with rewards 40, 20 

This problem can be solved using the concept of "GREEDY ALGORITHM".
Time Complexity - O(n log n)

We iterate over the tasks array once, sorting it in descending order of 
reward. Then we iterate over the sorted tasks array once more, scheduling 
each task if its deadline is greater than the current time.The greedy choice 
at each point is to either schedule the current task if its deadline is 
greater than the current time, or skip it otherwise.
"""

class Task:
 def __init__(self, id, deadline, reward):
 self.id = id
 self.deadline = deadline
 self.reward = reward

def max_tasks(tasks_info : list[tuple[int]]) -> int:
 """
 >>> max_tasks([(4, 20), (1, 10), (1, 40), (1, 30)])
 (2, [2, 0])
 >>> max_tasks([(1, 10), (2, 20), (3, 30), (2, 40)])
 (2, [3, 2])
 """
 tasks = [Task(i, d, p) for i, (d, p) in enumerate(tasks_info)]

 tasks.sort(key=lambda task: task.reward, reverse=True)

 schedule = []
 current_time = 0

 for task in tasks:
 if task.deadline > current_time:
 schedule.append(task.id)
 current_time += 1

 return len(schedule),schedule

if __name__ == "__main__":
 import doctest
 doctest.testmod()

 print(max_tasks([(4, 20), (1, 10), (1, 40), (1, 30)]))