Make Sum Divisible by P | Simplest Explanation | Full Dry Run | Leetcode 1590 | codestorywithMIK
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This is the 112th Video of our Playlist "Array 1D/2D : Popular Interview Problems" by codestorywithMIK
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In this video we will try to solve a good Array Cumulative Sum related Problem : Make Sum Divisible by P | Simplest Explanation | Full Dry Run | Leetcode 1590 | 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 : Make Sum Divisible by P | Simplest Explanation | Full Dry Run | Leetcode 1590 | codestorywithMIK
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My solutions on Github(C++ & JAVA) - https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/Arrays/Cumulative_Sum(Prefix Array)/Make Sum Divisible by P.cpp
Leetcode Link : https://leetcode.com/problems/make-sum-divisible-by-p
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Summary :
The solution aims to find the smallest subarray that, when removed, makes the sum of the remaining array divisible by a given integer p. Here’s how it works:
Calculate Total Modulo: First, compute the total sum of the array modulo p. This determines the target remainder that needs to be removed to make the remaining sum divisible by p.
Early Exit: If the total sum is already divisible by p (target == 0), the function immediately returns 0, as no subarray needs to be removed.
Iterate through the Array:
Maintain a prefix sum (curr) modulo p while iterating through the array.
Use a hash map (map) to store the indices of seen remainders to efficiently find subarrays with a specific remainder.
Check for Subarray:
For each element, calculate the remainder remain that, if removed, would make the remaining sum divisible by p.
If this remainder has been seen before, update the result with the length of the smallest subarray.
Return Result:
If a valid subarray is found, return its length. Otherwise, return -1.
This approach leverages prefix sums and hash maps for efficient subarray length calculation, making it run in linear time O(n).
✨ Timelines✨
00:00 - Introduction
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