Skip to content
Programming101
Programmingoneonone

Learn everything about programming

  • Home
  • CS Subjects
    • Internet of Things (IoT)
    • Digital Communication
    • Human Values
  • Programming Tutorials
    • C Programming
    • Data structures and Algorithms
    • 100+ Java Programs
    • 100+ C Programs
  • HackerRank Solutions
    • HackerRank Algorithms Solutions
    • HackerRank C problems solutions
    • HackerRank C++ problems solutions
    • HackerRank Java problems solutions
    • HackerRank Python problems solutions
Programming101
Programmingoneonone

Learn everything about programming

HackerRank Array Construction problem solution

YASH PAL, 31 July 2024

In this HackerRank Array Construction problem solution, we need to construct an n-element array A where the sum of all elements is equal to S and the sum of the absolute difference between each pair of elements is equal to k. All elements in A must be nonnegative integers.

HackerRank Array Construction problem solution

Problem solution in Python.

cumsum = []
res = []

def mn(ceil, beans):
    return (beans // ceil) * cumsum[ceil] + cumsum[beans % ceil]
def mx(ceil, beans):
    return cumsum[1] * beans

fmemo = set()
def f(ceil, sm, beans):
    if not (mn(ceil, beans) <= sm <= mx(ceil, beans)):
        return False
    if beans == 0 and sm == 0:
        return True
    if (ceil, sm, beans) in fmemo:
        return False
    
    for k in range(1, min(beans, ceil) + 1):
        if f(k, sm - cumsum[k], beans - k):
            res.append(k)
            return True
    fmemo.add((ceil, sm, beans))
    return False
    
        

for _ in range(int(input())):
    n, s, k = map(int, input().split())
    if n == 1:
        if k == 0:
            print(s)
        else:
            print(-1)
        continue
    if s == 0:
        if k == 0:
            print(' '.join(map(str,[0 for _ in range(n)])))
        else:
            print(-1)
        continue
    
    cumsum = [0, 2*n - 2]
    for i in range(1, n):
        cumsum.append(cumsum[-1] + 2*n - 2 - 2*i)
    res = []
    f(n, k + (n - 1) * s, s)
    if res:
        r = [0 for _ in range(n)]
        for num in res:
            for i in range(num):
                r[-1-i] += 1
        print(' '.join(map(str,r)))
    else:
        print(-1)

{“mode”:”full”,”isActive”:false}

Problem solution in Java.

import java.util.Arrays;
import java.util.Scanner;

public class Solution {
public static int[][][][] dp;
public static int[] construct(int nums, int sum1, int sum2) {
if (nums == 0) {
if (sum1 != 0 || sum2 != 0) {
return new int[]{-1};
}
return new int[]{};
}
if (dp[nums][sum1][sum2] != null) return dp[nums][sum1][sum2];

for (int place = 0; nums*place <= sum1; place++) {
int nsum1 = sum1 - place * nums;
int nsum2 = sum2 - place * nums - nsum1;
if (nsum2 < 0) continue;

int[] xx = construct(nums-1, nsum1, nsum2);
if (xx.length == 1 && xx[0] == -1) {
continue;
}
int[] actual = new int[nums];
System.arraycopy(xx, 0, actual, 1, xx.length);
for (int i = 0; i < nums; i++) actual[i] += place;
return dp[nums][sum1][sum2] = actual;
}
return dp[nums][sum1][sum2] = new int[]{-1};
}

public static void main (String[] args) {
dp = new int[51][201][5001][];
Scanner in = new Scanner(System.in);
int q = in.nextInt();
while(q-->0) {
int n = in.nextInt(), s = in.nextInt(), k = in.nextInt();
int[] min = null;
for (int start = 0; start*n <= s; start++) {
int[] b = construct(n-1, s-start*n, k);
if (b.length == 1 && b[0] == -1) {
continue;
}
int[] xx = new int[n];
System.arraycopy(b, 0, xx, 1, b.length);
for (int i = 0; i < n; i++) xx[i] += start;
if (min == null) {
min = xx;
continue;
}
boolean less = false;
for (int i = 0; i < n; i++) {
if (min[i] != xx[i]) {
if (xx[i] < min[i]) {
less = true;
}
break;
}
}
if (less) {
min = xx;
}
}

if (min != null) {
for (int i = 0; i < n; i++) {
if (i != 0) System.out.print(" ");
System.out.print(min[i]);
}
System.out.println();
} else {
System.out.println(-1);
}
}
System.exit(0);
}
}

{“mode”:”full”,”isActive”:false}

Problem solution in C++.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

const int Maxn = 52;
const int Maxs = 205;
const int Maxk = 2005;

bool dp[Maxn][Maxs][Maxk];

int main() {
    dp[0][0][0] = true;
    for (int i = 0; i < Maxn; i++)
        for (int j = 0; j < Maxs; j++)
            for (int k = 0; k < Maxk; k++) if (dp[i][j][k]) {
                if (j + i < Maxs) dp[i][j + i][k] = true;
                if (i + 1 < Maxn && k + j < Maxk) dp[i + 1][j][k + j] = true;
            }
    int q; scanf("%d", &q);
    while (q--) {
        int n, s, k; scanf("%d %d %d", &n, &s, &k);
        if (!dp[n][s][k]) { printf("-1n"); continue; }
        int el = 0;
        while (n > 0)
            if (k - s >= 0 && dp[n - 1][s][k - s]) {
                printf("%d%c", el, n - 1 > 0? ' ': 'n'); n--; k -= s;
            } else {
                el++;
                s -= n;
            }
    }
    return 0;
}

{“mode”:”full”,”isActive”:false}

Problem solution in C.

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#define N 50
#define S 200 
#define K 2000

int dp[N+10][S+10][K+10];
int a[N+10];
int main()
{
    int t;
    scanf("%d",&t);
    int n,s,k;
    for(int i=0; i<=N; i++)
    {
        for(int j=0; j<=S; j++)
        {
            for(int k=0; k<=K; k++)
            {
                dp[i][j][k]=-1;
            }
        }
    }
    dp[0][0][0] = 201;
    for(int i=1; i<=N; i++)
    {
        for(int j=0; j<=S; j++)
        {
            for(int k=0; k<=K; k++)
            {
                for(int l=0;l<=S/i; l++)
                {
                    if((k-j+i*l >= 0) && i*l <= j && j-l >=0 && dp[i-1][j-l][k-j+i*l]>=l)
                    {
                        dp[i][j][k] = l;
                    }
                }
            }
        }
    }
    while(t--)
    {
        scanf("%d%d%d",&n,&s,&k);
        if(dp[n][s][k] == -1)
        {
            printf("-1n");
        }
        else{
            int x = 0;
            for(int i=n; i>0; i--)
            {
                int f=0;
                for(int j=x; j<=s; j++)
                {
                    if(!f && k-s+i*j>=0 && dp[i-1][s-j][k-s+i*j]>=j)
                    {
                        k-=s-i*j;
                        s-=j;
                        x=j;
                        a[n-i+1]=j;
                        f=1;
                    }
                }
            }
            for(int i=1; i<=n; i++)
            {
                printf("%d ",a[i]);
            }
            printf("n");
        }
    }
    return 0;
}

{“mode”:”full”,”isActive”:false}

Algorithms coding problems solutions

Post navigation

Previous post
Next post

Pages

  • About US
  • Contact US
  • Privacy Policy

Programing Practice

  • C Programs
  • java Programs

HackerRank Solutions

  • C
  • C++
  • Java
  • Python
  • Algorithm

Other

  • Leetcode Solutions
  • Interview Preparation

Programming Tutorials

  • DSA
  • C

CS Subjects

  • Digital Communication
  • Human Values
  • Internet Of Things
  • YouTube
  • LinkedIn
  • Facebook
  • Pinterest
  • Instagram
©2025 Programmingoneonone | WordPress Theme by SuperbThemes