HackerEarth Minimize Cost problem solution

YASH PAL, 31 July 202413 February 2026

In this HackerEarth Minimize Cost problem solution, you are given an array of numbers Ai which contains positive as well as negative numbers. the cost of the array can be defined as C(X) = |A1 + T1 | + |A2 + T2| + ……|An + Tn|. where T is the transfer array which contains N zeros initially.

You need to minimize this cost. you can transfer value from one array element to another if and only if the distance between them is at most k.

hackerEarth Minimize Cost problem solution

HackerEarth Minimize Cost problem solution.

#include <iostream>
using namespace std;

int main()
{
    int n,k;
    freopen("input8.txt","r",stdin);
    freopen("output8.txt","w",stdout);
    cin>>n>>k;
    int arr[n];
    for(int i=0;i<n;i++)
    cin>>arr[i];
    int j = 0;
    for(int i=0;i<n;i++){
        
        if(arr[i]<0)
        continue;     
        
        while(i-j>k)
        ++j;
        
       
        while(arr[i]!=0 && (i+k)>=min(n-1,j)){
            if(arr[j]>0){
                j++;
                continue;
            }
            int x = min(arr[i],abs(arr[j]));
            arr[i]-=x;
            arr[j]+=x;
            if(arr[j]>=0)
                j++;
            }          
    }
    long long  ans =0;
    for(int i=0;i<n;i++)
    {
    ans+=abs(arr[i]);
    }
    cout<<ans;
}

second solution

#include<bits/stdc++.h>
#define LL long long int
#define M 1000000007
#define MM 1000000009
#define reset(a) memset(a,0,sizeof(a))
#define rep(i,j,k) for(i=j;i<=k;++i)
#define per(i,j,k) for(i=j;i>=k;--i)
#define print(a,start,end) for(i=start;i<=end;++i) cout<<a[i];
#define endl "n"
#define inf 100000000000000
LL pow(LL a,LL b,LL m){LL x=1,y=a;while(b > 0){if(b%2 == 1){x=(x*y);if(x>m) x%=m;}y = (y*y);if(y>m) y%=m;b /= 2;}return x%m;}
LL gcd(LL a,LL b){if(b==0) return a; else return gcd(b,a%b);}
LL gen(LL start,LL end){LL diff = end-start;LL temp = rand()%start;return temp+diff;}
using namespace std;
LL a[100001];
int main()
    {
        ios_base::sync_with_stdio(0);
        int n , k;
        cin >> n >> k;
        for(int i = 1; i <= n ; i++)
            cin >> a[i];
        int idx = 1;
        for(int i = 1; i <= n; i++)
            {
                again:;
                if(a[i] <= 0)
                    continue;
                while(a[idx] >= 0 && idx <= n && idx <= i + k)
                    {
                        ++idx;
                    }
                if(idx == n + 1)
                    break;
                if(idx == i + k + 1)
                    {
                        continue;
                    }
                if(abs(a[idx]) <= a[i])
                    {
                        a[i] += a[idx];
                        a[idx] = 0;
                    }
                else
                    {
                        a[idx] += a[i];
                        a[i] = 0;
                    }
                goto again;
            }
        LL ans = 0;
        for(int i = 1; i <= n ; i++)
            {
                ans = ans + (LL)abs(a[i]);
            }
        cout << ans;
    }

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