Partition Problem

#1

https://www.interviewbit.com/problems/partitions/ problem link…
This ques is totally going from over my head
can u explain following code logic where i commented
https://ide.codingblocks.com/s/478909

#2

hello @joshisanjay8467

cnt is holding how many times sum/3 occured till i (not including i).
so whener u find s==total sum/3 u increment its(cnt) value by 1.

now whenever ur s(prefix sum) = two_third of total sum
that means remaining part has sum =1/3 of total sum.
so all we need is how many times total sum/3 occured before current point.
we will consider it as left partition and current point as right parition.
that is why we are doing ans+=cnt;
if still u face any issue then check my implementation

int Solution::solve(int A, vector<int> &B){
    int sum=0,su=0,ways=0;
    for(auto k : B)
        sum+=k;
        if(sum%3)
        return 0;
    map<int,int> mp;
    for(int i=0;i<B.size()-1;i++){
        su+=B[i];
        
        if(sum-su==sum/3){
            ways+=mp[su-sum/3];
        }
        mp[su]++;
    }
    if(sum==0){
        ways=mp[0];
        ways=ways*(ways-1)/2;
    }
    return ways;
    
}
#3

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