Split Array Largest Sum | Leetcode 410 Solution | Searching and Sorting

NADOS also enables doubt support, career opportunities and contests besides free of charge content for learning. Question Statement:
1. Given an array nums which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays.
2. Write an algorithm to minimize the largest sum among these m subarrays.

Topic: #BinarySearch #leetcode410 #SearchingAndSorting

Used #DataStructure: #Array

#TimeComplexity: O(N*log(S)) , where N is total number of elements and S is array sum.

#SpaceComplexity: O(1)

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Linked Questions:

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#Array #BinarySearch #leetcode #SearchingAndSorting



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Happy Programming !!! Pep it up 😍🤩
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#pepcoding #code #coder #codinglife #programming #coding #java #freeresources #datastrucutres #pepcode #competitive #competitiveprogramming #softwareengineer #engineering #engineer

[RishinderRana]

May be it is the language that gets in our mind better, I couldn't find any english video with such a good explanation. The phrases used are very catchy and even non programmers can understand a hard problem like this by watching this video. Thanks a ton for the effort you did for making this tutorial. It'll help everyone (who understands hindi).

[devangshah]

Absolutely top notch! After quite a bit of searching, and head scratching to develop the intuition for the binary search solution...the development of intuition by the presenter is incredible!

[nikunjgupta]

This was the best possible explanation for this question. Keep up the good work!

[jaspreetsingh]

I just coded this leetcode hard in 5 minutes because of just Manisha ma'am, thank you so much pepcoding !!



code ->


class Solution {
public int splitArray(int[] nums, int m) {

int max = Integer.MIN_VALUE;
int sum = 0;

for(int a : nums){
sum+=a;
max = Math.max(a, max);

}

int low = max;
int high = sum;
int ans = 0;

while(low <= high){

int mid = low + (high - low) / 2;
if(isPossible(nums, mid, m)){

ans = mid;
high = mid - 1;
}
else{
low = mid + 1;
}
}

return ans;

}

public static boolean isPossible(int[] arr, int mid, int m){

int sum = 0 ; int count = 1;

for(int a : arr){

sum+=a;

if(sum > mid){

count++; sum = a;
}

}

return count <= m;
}
}

[Code_Solver]

This is the best explanation so far i have seen

[hemant_pandey]

At 10:36, the sum is still 14 (less than 15) and not 17.

[rajaryan]

God level explanation, samjh aa gya, thankyou

[syamsreemanojreddy]

Absolutely loved the explanation and approach.

[madanmohanpachouly]

Crisp and clear .. thanks for sharing.

[nitaraj]

Very nice explanation. Thanks to Pepcoding team !! You are indeed doing a great job.

[shakibsiddiqui]

Mam how can i access your question video search on channel but your videos is not showing

[sanjeevanrawat]

it was good, how you relate this problem, with pages allocation problem. appreciated

[thekumarpriyansh]

mam iss series me total kitne video aenge? Abhi tk sare sawal bohot ache hai.

[dietpepsi]

painter''s partition prblm bhi dal dijye portal pr

[rockykumar]

More questions kon same types
de please

[mohitgupta]

8
250 74 159 181 23 45 129 174
6
on this test case, your code gives 250 as ans, but keeping 250 as barrier, only 5 subarrays will formed ?
Even having this loophole, why all coding platform accepts this code?

[understand___]

I will be appreciate it if you could launch English Version