Maximum Number of Integers to Choose From a Range I | Simple | Leetcode 2554 | codestorywithMIK
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This is the 38th Video of our Playlist "Greedy : Popular Interview Problems" by codestorywithMIK
In this video we will try to solve a good greedy problem : Maximum Number of Integers to Choose From a Range I | Straight Forward Simple | Leetcode 2554 | codestorywithMIK
I will explain the intuition so easily that you will never forget and start seeing this as cakewalk EASYYY.
We will do live coding after explanation and see if we are able to pass all the test cases.
Also, please note that my Github solution link below contains both C++ as well as JAVA code.
Problem Name : Maximum Number of Integers to Choose From a Range I | Straight Forward Simple | Leetcode 2554 | codestorywithMIK
Company Tags : will update later
My solutions on Github(C++ & JAVA) - https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/Greedy/Maximum Number of Integers to Choose From a Range I.cpp
Leetcode Link : https://leetcode.com/problems/maximum-number-of-integers-to-choose-from-a-range-i
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Summary :
The goal is to find the maximum count of integers between 1 and n that can be included without exceeding a given maxSum while excluding the integers in a banned list.
Data Structure for Efficient Lookup:
A HashSet is used to store the banned numbers, enabling O(1) time complexity for checking if a number is banned.
Iterative Inclusion:
Iterate through all numbers from 1 to n. For each number:
Skip it if it is present in the banned set.
Check if adding the number to the current sum (sum) stays within the maxSum. If so:
Increment the count and update the sum.
Stop the process if the sum exceeds maxSum.
Greedy Approach:
The numbers are processed sequentially, ensuring the smallest possible numbers are added first to maximize the count.
This approach ensures optimal performance with a time complexity of O(n), as each number from 1 to n is processed once, and set lookups are O(1).
✨ Timelines✨
00:00 - Introduction
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