Check if it is possible to get given sum by taking one element from each row

Given a 2-D array of N rows and M columns and an integer K. The task is to find whether is it possible to take exactly one element from each row of the given array and make a total sum equal to K. 
Examples:

Input: N = 2, M = 10, K = 5
arr = {{4, 0, 15, 3, 2, 20, 10, 1, 5, 4}, 
      {4, 0, 10, 3, 2, 25, 4, 1, 5, 4}}
Output: YES
Explanation:
Take 2 from first row and 3 from second row.
2 + 3 = 5
So, we can make 5 by taking exactly one element
from each row.

Input:N = 3, M = 5, K = 5
arr = {{4, 3, 4, 5, 4}, 
       {2, 2, 3, 4, 3}, 
       {2, 1, 3, 3, 2}}
Output: NO

Approach: This problem can be solved using Dynamic Programming. 

1. We can make a 2-D binary array DP[][] of N rows and K columns. where DP[i][j] = 1 represents that we can make a sum equal to j by taking exactly one element from each row till i.  
2. So, we will iterate the array from i = [0, N], k = [0, K] and check if the current sum(k) exist or not. 
3. If the current sum exist then we will iterate over the column and update the array for every possible sum which is less than or equal to K. 
   

Below is the implementation of the above approach

C++

// C++ implementation to find
// whether is it possible to
// make sum equal to K
#include <bits/stdc++.h>
using namespace std;

// Function that prints whether is it
// possible to make sum equal to K
void PossibleSum(int n, int m,
              vector<vector<int> > v,
              int k)
{
    int dp[n + 1][k + 1] = { 0 };

    // Base case
    dp[0][0] = 1;

    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j <= k; j++)
        {
            // Condition if we can make
            // sum equal to current column
            // by using above rows
            if (dp[i][j] == 1)
            {
                // Iterate through current
                // column and check whether
                // we can make sum less than
                // or equal to k
                for (int d = 0; d < m; d++)
                {
                    if ((j + v[i][d]) <= k)
                    {
                        dp[i + 1][j + v[i][d]] = 1;
                    }
                }
            }
        }
    }

    // Printing whether is it
    // possible or not
    if (dp[n][k] == 1)
        cout << "YES\n";
    else
        cout << "NO\n";
}

// Driver Code
int main()
{
    int N = 2, M = 10, K = 5;

    vector<vector<int> > arr = { { 4, 0, 15, 3, 2,
                                  20, 10, 1, 5, 4 },
                                 { 4, 0, 10, 3, 2,
                                  25, 4, 1, 5, 4 } };

    PossibleSum(N, M, arr, K);

    return 0;
}

Java

// Java implementation to find
// whether is it possible to
// make sum equal to K
import java.io.*;
import java.util.*;

class GFG {

// Function that prints whether is it
// possible to make sum equal to K
static void PossibleSum(int n, int m,
                        int[][] v, int k)
{
    int[][] dp = new int[n + 1][k + 1];

    // Base case
    dp[0][0] = 1;

    for(int i = 0; i < n; i++)
    {
       for(int j = 0; j <= k; j++)
       {

          // Condition if we can make
          // sum equal to current column
          // by using above rows
          if (dp[i][j] == 1)
          {

              // Iterate through current
              // column and check whether
              // we can make sum less than
              // or equal to k
              for(int d = 0; d < m; d++)
              {
                 if ((j + v[i][d]) <= k)
                 {
                     dp[i + 1][j + v[i][d]] = 1;
                 }
              }
          }
       }
    }

    // Printing whether is it
    // possible or not
    if (dp[n][k] == 1)
        System.out.println("YES");
    else
        System.out.println("NO");
}

// Driver code
public static void main(String[] args)
{
    int N = 2, M = 10, K = 5;
    int[][] arr = new int[][]{ { 4, 0, 15, 3, 2,
                                20, 10, 1, 5, 4 },
                               { 4, 0, 10, 3, 2,
                                25, 4, 1, 5, 4 } };
    PossibleSum(N, M, arr, K);
}
}

// This code is contributed by coder001

Python3

# Python3 implementation to find
# whether is it possible to
# make sum equal to K

# Function that prints whether is it
# possible to make sum equal to K
def PossibleSum(n, m, v, k):

    dp = [[0] * (k + 1) for i in range(n + 1)]

    # Base case
    dp[0][0] = 1

    for i in range(n):
        for j in range(k + 1):

            # Condition if we can make
            # sum equal to current column
            # by using above rows
            if dp[i][j] == 1:

                # Iterate through current
                # column and check whether
                # we can make sum less than
                # or equal to k
                for d in range(m):
                    if (j + v[i][d]) <= k:
                        dp[i + 1][j + v[i][d]] = 1

    # Printing whether is it
    # possible or not
    if dp[n][k] == 1:
        print("YES")
    else:
        print("NO")

# Driver Code
N = 2
M = 10
K = 5
arr = [ [ 4, 0, 15, 3, 2,
         20, 10, 1, 5, 4 ],
        [ 4, 0, 10, 3, 2,
          25, 4, 1, 5, 4 ] ]

PossibleSum(N, M, arr, K)

# This code is contributed by divyamohan123

Javascript

<script>

// Javascript implementation to find
// whether is it possible to
// make sum equal to K

// Function that prints whether is it
// possible to make sum equal to K
function PossibleSum(n, m, v, k)
{
    var dp = Array.from(Array(n+1), ()=>Array(k+1).fill(0));

    // Base case
    dp[0][0] = 1;

    for (var i = 0; i < n; i++)
    {
        for (var j = 0; j <= k; j++)
        {
            // Condition if we can make
            // sum equal to current column
            // by using above rows
            if (dp[i][j] == 1)
            {
                // Iterate through current
                // column and check whether
                // we can make sum less than
                // or equal to k
                for (var d = 0; d < m; d++)
                {
                    if ((j + v[i][d]) <= k)
                    {
                        dp[i + 1][j + v[i][d]] = 1;
                    }
                }
            }
        }
    }

    // Printing whether is it
    // possible or not
    if (dp[n][k] == 1)
        document.write( "YES");
    else
        document.write( "NO");
}

// Driver Code
var N = 2, M = 10, K = 5;
var arr = [ [ 4, 0, 15, 3, 2,
                              20, 10, 1, 5, 4 ],
                             [ 4, 0, 10, 3, 2,
                              25, 4, 1, 5, 4 ] ];
PossibleSum(N, M, arr, K);

// This code is contributed by rutvik_56.
</script>

C#

// C# implementation to find
// whether is it possible to
// make sum equal to K
using System;
class GFG{

// Function that prints whether is it
// possible to make sum equal to K
static void PossibleSum(int n, int m,
                        int[,] v, int k)
{
    int[,] dp = new int[n + 1, k + 1];

    // Base case
    dp[0, 0] = 1;

    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j <= k; j++)
        {

            // Condition if we can make
            // sum equal to current column
            // by using above rows
            if (dp[i, j] == 1)
            {

                // Iterate through current
                // column and check whether
                // we can make sum less than
                // or equal to k
                for(int d = 0; d < m; d++)
                {
                    if ((j + v[i, d]) <= k)
                    {
                        dp[i + 1, j + v[i, d]] = 1;
                    }
                }
            }
        }
    }

    // Printing whether is it
    // possible or not
    if (dp[n, k] == 1)
        Console.WriteLine("YES");
    else
        Console.WriteLine("NO");
}

// Driver code
public static void Main(String[] args)
{
    int N = 2, M = 10, K = 5;
    int[,] arr = new int[,]{ { 4, 0, 15, 3, 2,
                              20, 10, 1, 5, 4 },
                             { 4, 0, 10, 3, 2,
                              25, 4, 1, 5, 4 } };
    PossibleSum(N, M, arr, K);
}
}

// This code is contributed by 29AjayKumar

Output:

YES

Time Complexity: O(N * M * K)
Auxiliary Space: O(N*K)

Efficient Approach : using array instead of 2d matrix to optimize space complexity
In previous code we can se that dp[i][j] is dependent upon dp[i+1] or dp[i] so we can assume that dp[i] is current row and dp[i] is next row.

Implementations Steps :

• Create two vectors curr and next each of size K+1 , where K is the sum of all elements of arr.
• Initialize them with 0.
• Set the Base Case.
• Now In previous code change dp[i] to curr and change dp[i+1] to next to keep track only of the two main rows.
• After every iteration update current row to next row to iterate further.

Implementation :

C++

#include <bits/stdc++.h>
using namespace std;

// Function that prints whether is it
// possible to make sum equal to K
void PossibleSum(int n, int m, vector<vector<int> > v,
                 int k)
{

    // initialize 2 vectors determine two rows of DP
    vector<int> curr(k + 1, 0);
    vector<int> next(k + 1, 0);

    // Base case
    curr[0] = 1;

    // loop to compute all subproblems
    for (int i = 0; i < n; i++) {
        for (int j = 0; j <= k; j++) {
            // Condition if we can make
            // sum equal to current column
            // by using above rows
            if (curr[j] == 1) {
                // Iterate through current
                // column and check whether
                // we can make sum less than
                // or equal to k
                for (int d = 0; d < m; d++) {
                    if ((j + v[i][d]) <= k) {
                        next[j + v[i][d]] = 1;
                    }
                }
            }
        }
        // assiging all values of next to curr vector
        swap(curr, next);
        fill(next.begin(), next.end(), 0);
    }

    // Printing whether is it
    // possible or not
    if (curr[k] == 1)
        cout << "YES\n";
    else
        cout << "NO\n";
}

// Driver Code
int main()
{
    int N = 2, M = 10, K = 5;
    vector<vector<int> > arr
        = { { 4, 0, 15, 3, 2, 20, 10, 1, 5, 4 },
            { 4, 0, 10, 3, 2, 25, 4, 1, 5, 4 } };

    PossibleSum(N, M, arr, K);

    return 0;
}

Java

import java.util.*;

public class PossibleSum {

    // Function that prints whether it is
    // possible to make sum equal to K
    public static void
    possibleSum(int n, int m, List<List<Integer> > v, int k)
    {

        // initialize 2 vectors determine two rows of DP
        int[] curr = new int[k + 1];
        int[] next = new int[k + 1];

        // Base case
        curr[0] = 1;

        // loop to compute all subproblems
        for (int i = 0; i < n; i++) {
            for (int j = 0; j <= k; j++) {

                // Condition if we can make
                // sum equal to current column
                // by using above rows
                if (curr[j] == 1) {

                    // Iterate through current
                    // column and check whether
                    // we can make sum less than
                    // or equal to k
                    for (int d = 0; d < m; d++) {
                        if ((j + v.get(i).get(d)) <= k) {
                            next[j + v.get(i).get(d)] = 1;
                        }
                    }
                }
            }

            // assiging all values of next to curr vector
            curr = Arrays.copyOf(next, k + 1);
            Arrays.fill(next, 0);
        }

        // Printing whether it is possible or not
        if (curr[k] == 1)
            System.out.println("YES");
        else
            System.out.println("NO");
    }

    // Driver Code
    public static void main(String[] args)
    {
        int N = 2, M = 10, K = 5;
        List<List<Integer> > arr = new ArrayList<>();
        arr.add(
            Arrays.asList(4, 0, 15, 3, 2, 20, 10, 1, 5, 4));
        arr.add(
            Arrays.asList(4, 0, 10, 3, 2, 25, 4, 1, 5, 4));

        possibleSum(N, M, arr, K);
    }
}

Python3

from typing import List


def possibleSum(n: int, m: int, v: List[List[int]], k: int) -> None:
    # initialize 2 vectors determine two rows of DP
    curr = [0] * (k + 1)
    next = [0] * (k + 1)

    # Base case
    curr[0] = 1

    # loop to compute all subproblems
    for i in range(n):
        for j in range(k+1):
            # Condition if we can make
            # sum equal to current column
            # by using above rows
            if curr[j] == 1:
                # Iterate through current
                # column and check whether
                # we can make sum less than
                # or equal to k
                for d in range(m):
                    if j + v[i][d] <= k:
                        next[j + v[i][d]] = 1

        # assiging all values of next to curr vector
        curr = next.copy()
        next = [0] * (k + 1)

    # Printing whether it is possible or not
    if curr[k] == 1:
        print("YES")
    else:
        print("NO")


# Driver Code
if __name__ == '__main__':
    N, M, K = 2, 10, 5
    arr = [
        [4, 0, 15, 3, 2, 20, 10, 1, 5, 4],
        [4, 0, 10, 3, 2, 25, 4, 1, 5, 4]
    ]
    possibleSum(N, M, arr, K)

C#

using System;
using System.Collections.Generic;

public class Program {
    public static void
    PossibleSum(int n, int m, List<List<int> > v, int k)
    {
        // initialize 2 vectors determine two rows of DP
        List<int> curr = new List<int>(new int[k + 1]);
        List<int> next = new List<int>(new int[k + 1]);

        // Base case
        curr[0] = 1;

        // loop to compute all subproblems
        for (int i = 0; i < n; i++) {
            for (int j = 0; j <= k; j++) {
                // Condition if we can make
                // sum equal to current column
                // by using above rows
                if (curr[j] == 1) {
                    // Iterate through current
                    // column and check whether
                    // we can make sum less than
                    // or equal to k
                    for (int d = 0; d < m; d++) {
                        if ((j + v[i][d]) <= k) {
                            next[j + v[i][d]] = 1;
                        }
                    }
                }
            }
            // assiging all values of next to curr vector
            curr = new List<int>(next);
            next = new List<int>(new int[k + 1]);
        }

        // Printing whether is it
        // possible or not
        if (curr[k] == 1)
            Console.WriteLine("YES");
        else
            Console.WriteLine("NO");
    }

    // Driver Code
    public static void Main()
    {
        int N = 2, M = 10, K = 5;
        List<List<int> > arr = new List<List<int> >{
            new List<int>{ 4, 0, 15, 3, 2, 20, 10, 1, 5,
                           4 },
            new List<int>{ 4, 0, 10, 3, 2, 25, 4, 1, 5, 4 }
        };

        PossibleSum(N, M, arr, K);
    }
}

Javascript

function possibleSum(n, m, v, k) {
  // initialize 2 arrays determine two rows of DP
  let curr = Array(k + 1).fill(0);
  let next = Array(k + 1).fill(0);

  // Base case
  curr[0] = 1;

  // loop to compute all subproblems
  for (let i = 0; i < n; i++) {
    for (let j = 0; j <= k; j++) {
      // Condition if we can make
      // sum equal to current column
      // by using above rows
      if (curr[j] === 1) {
        // Iterate through current
        // column and check whether
        // we can make sum less than
        // or equal to k
        for (let d = 0; d < m; d++) {
          if (j + v[i][d] <= k) {
            next[j + v[i][d]] = 1;
          }
        }
      }
    }
    // assiging all values of next to curr array
    curr = [...next];
    next.fill(0);
  }

  // Printing whether it is possible or not
  if (curr[k] === 1) {
    console.log("YES");
  } else {
    console.log("NO");
  }
}

// Driver Code
const n = 2, m = 10, k = 5;
const arr = [
  [4, 0, 15, 3, 2, 20, 10, 1, 5, 4],
  [4, 0, 10, 3, 2, 25, 4, 1, 5, 4]
];
possibleSum(n, m, arr, k);

Output

YES

Time Complexity: O(N * K)
Auxiliary Space: O(K)