1769. Minimum Number of Operations to Move All Balls to Each Box | leetcode daily challenge | dsa

Problem Link:
leetcode.com/p...
Problem Statement:
You have n boxes. You are given a binary string boxes of length n, where boxes[i] is '0' if the ith box is empty, and '1' if it contains one ball.
In one operation, you can move one ball from a box to an adjacent box. Box i is adjacent to box j if abs(i - j) == 1. Note that after doing so, there may be more than one ball in some boxes.
Return an array answer of size n, where answer[i] is the minimum number of operations needed to move all the balls to the ith box.
Each answer[i] is calculated considering the initial state of the boxes.
Solution Link:
github.com/Tiw...
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Notes:
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Timestamp:
0:00 - Introduction

[@shashwat_tiwari_st]

Like target is 150. Please do like if you have understood the explanation as well as the code ❤

[@chandanpandey1847]

You deserve more ❤

[@Rahul_Mongia]

similar to product of array except itself , wonderful explanation sir 💜

[@shashwat_tiwari_st]

@@Rahul_Mongia 100%

[@GautamParmar-ug9ld]

Bhiaya leetcode weekly contest pr bhi video banao please ......it would be really helpfull for us

[@OnstreamGaming]

bnda job krta hai , full time youtiube nhi

[@OnstreamGaming]

bruiteforce done

[@shivendrasinghthakur4476]

There is no use for the prevRight variable.

[@nishith5380]

Bro how will you evn think like this .great

[@SAI-q1v]

Bhaiya optimal idea nahi aaraha he brute hohi ja raha

[@shashwat_tiwari_st]

@@SAI-q1v product of array except itself solve kra tha pehle? Same intuition h

[@SAI-q1v]

@shashwat_tiwari_st nahi bhaiya karunga❤️

[@srishanth-r7n]

public int[] minOperations(String boxes) {
int n=boxes.length();
int res[]=new int[n];
for(int i=0;i

[@asad_iliyas]

It was easy brute force

[@dayashankarlakhotia4943]

public int[]minOperations(String boxes){
int n=boxes.length();
int[]ans=new int[n];
for(int i=0,ops=0,cnt=0;i=0;i--){
ans[i]+=ops;
cnt+=boxes.charAt(i)-'0';
ops+=cnt;
}
return ans;
tc=0(n);
sc=0(n);🎉❤

[@Ajay-cv1zs]

More simple solution (golang)
func minOperations(boxes string) []int {
n := len(boxes)
result:=make([]int,n)
balls:=0
moves:=0
for i:=0;i=0;i--{
result[i]+=moves
if boxes[i]=='1'{
balls++
}
moves+=balls
}
return result
}