Longest Common Subsequence with no repeating character

Length of Longest Common Subsequence with given sum K

Last Updated : 03 Apr, 2023

Given two arrays a[] and b[] and an integer K, the task is to find the length of the longest common subsequence such that sum of elements is equal to K.

Examples:

Input: a[] = { 9, 11, 2, 1, 6, 2, 7}, b[] = {1, 2, 6, 9, 2, 3, 11, 7}, K = 18
Output: 3
Explanation: Subsequence { 11, 7 } and { 9, 2, 7 } has sum equal to 18.
Among them { 9, 2, 7 } is longest. Therefore the output will be 3.

Input: a[] = { 2, 5, 2, 5, 7, 9, 4, 2}, b[] = { 1, 6, 2, 7, 8 }, K = 8
Output: -1

Approach: The approach to the solution is based on the concept of longest common subsequence and we need to check if sum of elements of subsequence is equal to given value. Follow the steps mentioned below;

Consider variable sum initialized to given value.
Each time when elements are included in subsequence, decrease sum value by this element.
In base condition check if sum value is 0 which implies this subsequence has sum equal to K. Therefore return 0 when sum is zero, else return INT_MIN

Below is the implementation of the above approach :

C++

// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
int solve(int a[], int b[], int i, int j, int sum)
{
    if (sum == 0)
        return 0;
    if (sum < 0)
        return INT_MIN;

    if (i == 0 || j == 0) {
        if (sum == 0)
            return 0;
        else
            return INT_MIN;
    }

    // If values are same then we can include this
    // or also can't include this
    if (a[i - 1] == b[j - 1])
        return max(
            1 + solve(a, b, i - 1, j - 1, sum - a[i - 1]),
            solve(a, b, i - 1, j - 1, sum));

    return max(solve(a, b, i - 1, j, sum),
               solve(a, b, i, j - 1, sum));
}

// Driver code
int main()
{
    int a[] = { 9, 11, 2, 1, 6, 2, 7 };
    int b[] = { 1, 2, 6, 9, 2, 3, 11, 7 };
    int n = sizeof(a) / sizeof(int);
    int m = sizeof(b) / sizeof(int);
    int sum = 18;

    int ans = solve(a, b, n, m, sum);
    if (ans >= 0)
        cout << ans << endl;
    else
        cout << -1;
    return 0;
}

Java

// Java code to implement the approach
import java.io.*;
class GFG {

  static int solve(int a[], int b[], int i, int j, int sum)
  {
    if (sum == 0)
      return 0;
    if (sum < 0)
      return Integer.MIN_VALUE;

    if (i == 0 || j == 0) {
      if (sum == 0)
        return 0;
      else
        return Integer.MIN_VALUE;
    }

    // If values are same then we can include this
    // or also can't include this
    if (a[i - 1] == b[j - 1])
      return Math.max(
      1 + solve(a, b, i - 1, j - 1, sum - a[i - 1]),
      solve(a, b, i - 1, j - 1, sum));

    return Math.max(solve(a, b, i - 1, j, sum),
                    solve(a, b, i, j - 1, sum));
  }

  // Driver code
  public static void main (String[] args) {
    int a[] = { 9, 11, 2, 1, 6, 2, 7 };
    int b[] = { 1, 2, 6, 9, 2, 3, 11, 7 };
    int n = a.length;
    int m = b.length;
    int sum = 18;

    int ans = solve(a, b, n, m, sum);
    if (ans >= 0)
      System.out.print(ans);
    else
      System.out.print(-1);
  }
}

// This code is contributed by hrithikgarg03188.

Python3

# Python code for the above approach
def solve(a, b,  i,  j,  sum):

    if sum == 0:
        return 0
    if sum < 0:
        return -2147483648

    if i == 0 or j == 0:
        if sum == 0:
            return 0
        else:
            return -2147483648

    # If values are same then we can include this
    # or also can't include this
    if (a[i - 1] == b[j - 1]):
        return max(
            1 + solve(a, b, i - 1, j - 1, sum - a[i - 1]),
            solve(a, b, i - 1, j - 1, sum))

    return max(solve(a, b, i - 1, j, sum),
               solve(a, b, i, j - 1, sum))

# Driver code
a = [9, 11, 2, 1, 6, 2, 7]
b = [1, 2, 6, 9, 2, 3, 11, 7]
n = len(a)
m = len(b)
sum = 18

ans = solve(a, b, n, m, sum)
if (ans >= 0):
    print(ans)
else:
    print(-1)

# This code is contributed by Potta Lokesh

// C# code to implement the approach
using System;
class GFG {

  static int solve(int[] a, int[] b, int i, int j,
                   int sum)
  {
    if (sum == 0)
      return 0;
    if (sum < 0)
      return Int32.MinValue;

    if (i == 0 || j == 0) {
      if (sum == 0)
        return 0;
      else
        return Int32.MinValue;
    }

    // If values are same then we can include this
    // or also can't include this
    if (a[i - 1] == b[j - 1])
      return Math.Max(1
                      + solve(a, b, i - 1, j - 1,
                              sum - a[i - 1]),
                      solve(a, b, i - 1, j - 1, sum));

    return Math.Max(solve(a, b, i - 1, j, sum),
                    solve(a, b, i, j - 1, sum));
  }

  // Driver code
  public static void Main()
  {
    int[] a = { 9, 11, 2, 1, 6, 2, 7 };
    int[] b = { 1, 2, 6, 9, 2, 3, 11, 7 };
    int n = a.Length;
    int m = b.Length;
    int sum = 18;

    int ans = solve(a, b, n, m, sum);
    if (ans >= 0)
      Console.Write(ans);
    else
      Console.Write(-1);
  }
}

// This code is contributed by Samim Hossain Mondal..

JavaScript

    <script>
        // JavaScript code to implement the approach
        const INT_MIN = -2147483648;

        const solve = (a, b, i, j, sum) => {
            if (sum == 0)
                return 0;
            if (sum < 0)
                return INT_MIN;

            if (i == 0 || j == 0) {
                if (sum == 0)
                    return 0;
                else
                    return INT_MIN;
            }

            // If values are same then we can include this
            // or also can't include this
            if (a[i - 1] == b[j - 1])
                return Math.max(
                    1 + solve(a, b, i - 1, j - 1, sum - a[i - 1]),
                    solve(a, b, i - 1, j - 1, sum));

            return Math.max(solve(a, b, i - 1, j, sum),
                solve(a, b, i, j - 1, sum));
        }

        // Driver code

        let a = [9, 11, 2, 1, 6, 2, 7];
        let b = [1, 2, 6, 9, 2, 3, 11, 7];
        let n = a.length;
        let m = b.length;
        let sum = 18;

        let ans = solve(a, b, n, m, sum);
        if (ans >= 0)
            document.write(`${ans}`);
        else
            document.write("-1");

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

Output

Time Complexity: O(2^N), N max size among both array
Auxiliary Space: O(1)

Efficient approach: An efficient approach is to use memoization to reduce the time complexity where each state stores the maximum length of a subsequence having a sum. Use map to achieve this.

Below is the implementation of the above approach.

C++

// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;

// Function to find longest common subsequence having sum
// equal to given value
int solve(int a[], int b[], int i, int j, int sum,
          map<string, int>& mp)
{
    if (sum == 0)
        return 0;

    if (sum < 0)
        return INT_MIN;

    if (i == 0 || j == 0) {
        if (sum == 0)
            return 0;
        else
            return INT_MIN;
    }

    string temp = to_string(i) + '#'
                  + to_string(j) + '#'
                  + to_string(sum);
    if (mp.find(temp) != mp.end())
        return mp[temp];

    // If values are same then we can include this
    // or also can't include this
    if (a[i - 1] == b[j - 1])
        return mp[temp]
               = max(
                   1 + solve(a, b, i - 1, j - 1,
                             sum - a[i - 1], mp),
                   solve(a, b, i - 1, j - 1, sum, mp));

    return mp[temp]
           = max(solve(a, b, i - 1, j, sum, mp),
                 solve(a, b, i, j - 1, sum, mp));
}

// Driver code
int main()
{
    int a[] = { 9, 11, 2, 1, 6, 2, 7 };
    int b[] = { 1, 2, 6, 9, 2, 3, 11, 7 };
    map<string, int> mp;
    int n = sizeof(a) / sizeof(int);
    int m = sizeof(b) / sizeof(int);
    int sum = 18;

    int ans = solve(a, b, n, m, sum, mp);
    if (ans >= 0)
        cout << ans << endl;
    else
        cout << -1;
    return 0;
}

Java

// Java code to implement the approach
import java.util.*;

class GFG{

  // Function to find longest common subsequence having sum
  // equal to given value
  static int solve(int a[], int b[], int i, int j, int sum,
                   HashMap<String, Integer> mp)
  {
    if (sum == 0)
      return 0;

    if (sum < 0)
      return Integer.MIN_VALUE;

    if (i == 0 || j == 0) {
      if (sum == 0)
        return 0;
      else
        return Integer.MIN_VALUE;
    }

    String temp = String.valueOf(i) + '#'
      + String.valueOf(j) + '#'
      + String.valueOf(sum);
    if (mp.containsKey(temp))
      return mp.get(temp);

    // If values are same then we can include this
    // or also can't include this
    if (a[i - 1] == b[j - 1]) {
      mp.put(temp, Math.max(
        1 + solve(a, b, i - 1, j - 1,
                  sum - a[i - 1], mp),
        solve(a, b, i - 1, j - 1, sum, mp)));
      return mp.get(temp);
    }

    mp.put(temp,  Math.max(solve(a, b, i - 1, j, sum, mp),
                           solve(a, b, i, j - 1, sum, mp)));
    return mp.get(temp);
  }

  // Driver code
  public static void main(String[] args)
  {
    int a[] = { 9, 11, 2, 1, 6, 2, 7 };
    int b[] = { 1, 2, 6, 9, 2, 3, 11, 7 };
    HashMap<String, Integer> mp = new HashMap<>();
    int n = a.length;
    int m = b.length;
    int sum = 18;

    int ans = solve(a, b, n, m, sum, mp);
    if (ans >= 0)
      System.out.print(ans +"\n");
    else
      System.out.print(-1);
  }
}

// This code is contributed by shikhasingrajput

Python3

# Python code to implement the approach

# Function to find longest common subsequence having sum
# equal to given value
def solve(a, b, i, j, sum, mp):

    if sum == 0:
        return 0
    if sum < 0:
        return -2147483648

    if i == 0 or j == 0:
        if sum == 0:
            return 0
        else:
            return -2147483648
    
    temp=str(i)+"#"+str(j)+"#"+str(sum)
    if(temp in mp):
        return mp[temp]
    
    # If values are same then we can include this
    # or also can't include this
    if (a[i - 1] == b[j - 1]):
        mp[temp] = max(1 + solve(a, b, i - 1, j - 1, sum - a[i - 1], mp),solve(a, b, i - 1, j - 1, sum,mp))
        return mp[temp]
        
    mp[temp] = max(solve(a, b, i - 1, j, sum,mp),solve(a, b, i, j - 1, sum,mp))
    return mp[temp]
    
# Driver code
a = [9, 11, 2, 1, 6, 2, 7]
b = [1, 2, 6, 9, 2, 3, 11, 7]
n = len(a)
m = len(b)
sum = 18
mp = {}
ans = solve(a, b, n, m, sum, mp)
if (ans >= 0):
    print(ans)
else:
    print(-1)

# This code is contributed by Pushpesh Raj

// C# code to implement the approach
using System;
using System.Collections.Generic;
class GFG {

  // Function to find longest common subsequence having
  // sum equal to given value
  static int solve(int[] a, int[] b, int i, int j,
                   int sum, Dictionary<string, int> mp)
  {
    if (sum == 0)
      return 0;

    if (sum < 0)
      return Int32.MinValue;

    if (i == 0 || j == 0) {
      if (sum == 0)
        return 0;
      else
        return Int32.MinValue;
    }

    string temp = i.ToString() + "#" + j.ToString()
      + "#" + sum.ToString();
    if (mp.ContainsKey(temp))
      return mp[temp];

    // If values are same then we can include this
    // or also can't include this
    if (a[i - 1] == b[j - 1])
      return mp[temp] = Math.Max(
      1
      + solve(a, b, i - 1, j - 1,
              sum - a[i - 1], mp),
      solve(a, b, i - 1, j - 1, sum, mp));

    return mp[temp]
      = Math.Max(solve(a, b, i - 1, j, sum, mp),
                 solve(a, b, i, j - 1, sum, mp));
  }

  // Driver code
  public static void Main()
  {
    int[] a = { 9, 11, 2, 1, 6, 2, 7 };
    int[] b = { 1, 2, 6, 9, 2, 3, 11, 7 };
    Dictionary<string, int> mp
      = new Dictionary<string, int>();

    int n = a.Length;
    int m = b.Length;
    int sum = 18;

    int ans = solve(a, b, n, m, sum, mp);
    if (ans >= 0)
      Console.WriteLine(ans);
    else
      Console.Write(-1);
  }
}

// This code is contributed by Samim Hossain Mondal.

JavaScript

// JavaScript code to implement the approach

// Function to find longest common subsequence having sum
// equal to given value
function solve(a, b, i, j, sum, mp) {

if (sum == 0) {
return 0;
}
if (sum < 0) {
return -2147483648;
}

if (i == 0 || j == 0) {
if (sum == 0) {
return 0;
} else {
return -2147483648;
}
}

const temp = i + "#" + j + "#" + sum;
if (temp in mp) {
return mp[temp];
}

// If values are same then we can include this
// or also can't include this
if (a[i - 1] == b[j - 1]) {
mp[temp] = Math.max(1 + solve(a, b, i - 1, j - 1, sum - a[i - 1], mp), 
solve(a, b, i - 1, j - 1, sum, mp));
return mp[temp];
}

mp[temp] = Math.max(solve(a, b, i - 1, j, sum, mp), 
solve(a, b, i, j - 1, sum, mp));
return mp[temp];
}

// Driver code
const a = [9, 11, 2, 1, 6, 2, 7];
const b = [1, 2, 6, 9, 2, 3, 11, 7];
const n = a.length;
const m = b.length;
const sum = 18;
const mp = {};
const ans = solve(a, b, n, m, sum, mp);
if (ans >= 0) {
console.log(ans);
} else {
console.log(-1);
}

Output

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

Another Approach : Using Dp tabulation (Iterative approach)

The approach to solve the problem is same but in this approach we solve the problem without recursion.

Implementation Steps :

Initialize a 3D DP array dp[n+1][m+1][sum+1] with all values set to 0.
Iterate over the indices i and j from 0 to n and m, respectively, and the sum k from 0 to sum.
For each index (i,j,k), check the following cases:
a. If k is 0, set dp[i][j][k] to 0.
b. If either i or j is 0 and k is not 0, set dp[i][j][k] to a very small negative value to indicate impossibility.
c. If the current elements of both arrays a and b are the same, set dp[i][j][k] to the maximum of either including or excluding the current element.
d. If the current elements of both arrays are different, set dp[i][j][k] to the maximum of the longest subsequence without the current element.
The final answer will be stored in dp[n][m][sum]. If the value is negative, the problem is impossible and the answer is -1. Otherwise, return the value of dp[n][m][sum].

below is the implementation of above approach :

C++

// C++ program for above approach

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

// DP array to store the lengths of longest common subsequences with a given sum
int dp[101][101][1001]; 

int solve(int a[], int b[], int n, int m, int sum) {
    memset(dp, 0, sizeof(dp)); // Initialize the DP array to 0
    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= m; j++) {
            for (int k = 0; k <= sum; k++) {
                 // Base case 1: sum is 0
                if (k == 0)
                    dp[i][j][k] = 0;
                // Base case 2: either array is empty    
                else if (i == 0 || j == 0) {
                    if (k == 0)
                        dp[i][j][k] = 0;
                    else
                        // Set to a very small negative value to indicate impossibility
                        dp[i][j][k] = -1e9; 
                }
                else if (a[i - 1] == b[j - 1]) { 
                    // Case 1: current elements of both arrays are the same
                    // Choose or don't choose the current element
                    dp[i][j][k] = max(1 + dp[i - 1][j - 1][k - a[i - 1]], dp[i - 1][j - 1][k]); 
                }
                else { // Case 2: current elements of both arrays are different
                    // Choose the longest subsequence without the current element
                    dp[i][j][k] = max(dp[i - 1][j][k], dp[i][j - 1][k]); 
                }
            }
        }
    }
    // Return the final answer, or -1 if it is impossible
    return (dp[n][m][sum] < 0) ? -1 : dp[n][m][sum]; 
}

// Driver code
int main() {
    int a[] = {9, 11, 2, 1, 6, 2, 7};
    int b[] = {1, 2, 6, 9, 2, 3, 11, 7};
    int n = sizeof(a) / sizeof(int);
    int m = sizeof(b) / sizeof(int);
    int sum = 18;
    
    // function call
    int ans = solve(a, b, n, m, sum);
    cout << ((ans == -1) ? -1 : ans) << endl; // Print the answer, or -1 if it is impossible
    return 0;
}

// this code is contributed by bhardwajji

Java

import java.util.Arrays;

public class Main 
{
  
  // DP array to store the lengths of longest 
  // common subsequences with a given sum
  static int[][][] dp = new int[101][101][1001];

  static int solve(int[] a, int[] b, int n, int m, int sum)
  {
    
    // Initialize the DP array to 0
    for (int[][] row : dp)
      for (int[] col : row)
        Arrays.fill(col, 0);

    for (int i = 0; i <= n; i++) {
      for (int j = 0; j <= m; j++) {
        for (int k = 0; k <= sum; k++) {
          // Base case 1: sum is 0
          if (k == 0)
            dp[i][j][k] = 0;
          // Base case 2: either array is empty
          else if (i == 0 || j == 0) {
            if (k == 0)
              dp[i][j][k] = 0;
            else
              // Set to a very small negative 
              // value to indicate impossibility
              dp[i][j][k] = -1000000000;
          } else if (a[i - 1] == b[j - 1]) {
            // Case 1: current elements of both arrays are the same
            // Choose or don't choose the current element
            if (k >= a[i - 1])
              dp[i][j][k] = Math.max(1 + dp[i - 1][j - 1][k - a[i - 1]], dp[i - 1][j - 1][k]);
            else
              dp[i][j][k] = dp[i - 1][j - 1][k];
          } else { // Case 2: current elements of both arrays are different
            // Choose the longest subsequence without the current element
            dp[i][j][k] = Math.max(dp[i - 1][j][k], dp[i][j - 1][k]);
          }
        }
      }
    }
    // Return the final answer, or -1 if it is impossible
    return (dp[n][m][sum] < 0) ? -1 : dp[n][m][sum];
  }

  public static void main(String[] args) {
    int[] a = {9, 11, 2, 1, 6, 2, 7};
    int[] b = {1, 2, 6, 9, 2, 3, 11, 7};
    int n = a.length;
    int m = b.length;
    int sum =18;

    // function call
    int ans = solve(a, b, n, m,sum);
    System.out.println((ans == -1) ? -1 : ans); // Print the answer or -1 if it is impossible
  }
}

Python3

# Python program for above approach

# DP array to store the lengths of longest common subsequences with a given sum
dp = [[[0 for _ in range(1001)] for _ in range(101)] for _ in range(101)]

def solve(a, b, n, m, sum):
    global dp
    # Initialize the DP array to 0
    for i in range(n+1):
        for j in range(m+1):
            for k in range(sum+1):
                # Base case 1: sum is 0
                if k == 0:
                    dp[i][j][k] = 0
                # Base case 2: either array is empty
                elif i == 0 or j == 0:
                    if k == 0:
                        dp[i][j][k] = 0
                    else:
                        # Set to a very small negative value to indicate impossibility
                        dp[i][j][k] = -1e9
                elif a[i-1] == b[j-1]:
                    # Case 1: current elements of both arrays are the same
                    # Choose or don't choose the current element
                    dp[i][j][k] = max(1 + dp[i-1][j-1][k-a[i-1]], dp[i-1][j-1][k])
                else:
                    # Case 2: current elements of both arrays are different
                    # Choose the longest subsequence without the current element
                    dp[i][j][k] = max(dp[i-1][j][k], dp[i][j-1][k])
    
    # Return the final answer, or -1 if it is impossible
    return -1 if dp[n][m][sum] < 0 else dp[n][m][sum]

# Driver code
if __name__ == "__main__":
    a = [9, 11, 2, 1, 6, 2, 7]
    b = [1, 2, 6, 9, 2, 3, 11, 7]
    n = len(a)
    m = len(b)
    sum = 18

    # function call
    ans = solve(a, b, n, m, sum)
    print(ans if ans != -1 else -1) # Print the answer, or -1 if it is impossible

using System;

public class GFG {
    // DP array to store the lengths of longest
    // common subsequences with a given sum
    static int[, , ] dp = new int[101, 101, 1001];

    static int solve(int[] a, int[] b, int n, int m,
                     int sum)
    {
        // Initialize the DP array to 0
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= m; j++) {
                for (int k = 0; k <= sum; k++) {
                    dp[i, j, k] = 0;
                }
            }
        }

        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= m; j++) {
                for (int k = 0; k <= sum; k++) {
                    // Base case 1: sum is 0
                    if (k == 0)
                        dp[i, j, k] = 0;
                    // Base case 2: either array is empty
                    else if (i == 0 || j == 0) {
                        if (k == 0)
                            dp[i, j, k] = 0;
                        else
                            // Set to a very small negative
                            // value to indicate
                            // impossibility
                            dp[i, j, k] = -1000000000;
                    }
                    else if (a[i - 1] == b[j - 1]) {
                        // Case 1: current elements of both
                        // arrays are the same Choose or
                        // don't choose the current element
                        if (k >= a[i - 1])
                            dp[i, j, k] = Math.Max(
                                1
                                    + dp[i - 1, j - 1,
                                         k - a[i - 1]],
                                dp[i - 1, j - 1, k]);
                        else
                            dp[i, j, k]
                                = dp[i - 1, j - 1, k];
                    }
                    else { // Case 2: current elements of
                           // both arrays are different
                        // Choose the longest subsequence
                        // without the current element
                        dp[i, j, k]
                            = Math.Max(dp[i - 1, j, k],
                                       dp[i, j - 1, k]);
                    }
                }
            }
        }
        // Return the final answer, or -1 if it is
        // impossible
        return (dp[n, m, sum] < 0) ? -1 : dp[n, m, sum];
    }

    public static void Main()
    {
        int[] a = { 9, 11, 2, 1, 6, 2, 7 };
        int[] b = { 1, 2, 6, 9, 2, 3, 11, 7 };
        int n = a.Length;
        int m = b.Length;
        int sum = 18;

        // function call
        int ans = solve(a, b, n, m, sum);
        Console.WriteLine(
            (ans == -1) ? -1
                        : ans); // Print the answer or -1 if
                                // it is impossible
    }
}

JavaScript

function solve(a, b, n, m, sum) {
  // DP array to store the lengths of longest common 
  // subsequences with a given sum
  const dp = new Array(n + 1)
    .fill()
    .map(() =>
      new Array(m + 1)
        .fill()
        .map(() => new Array(sum + 1).fill(0))
    );

  for (let i = 0; i <= n; i++) {
    for (let j = 0; j <= m; j++) {
      for (let k = 0; k <= sum; k++) {
        // Base case 1: sum is 0
        if (k === 0) {
          dp[i][j][k] = 0;
        }
        // Base case 2: either array is empty
        else if (i === 0 || j === 0) {
          if (k === 0) {
            dp[i][j][k] = 0;
          } else {
            // Set to a very small negative value to indicate impossibility
            dp[i][j][k] = -1000000000;
          }
        } else if (a[i - 1] === b[j - 1]) {
          // Case 1: current elements of both arrays are the same
          // Choose or don't choose the current element
          if (k >= a[i - 1]) {
            dp[i][j][k] = Math.max(
              1 + dp[i - 1][j - 1][k - a[i - 1]],
              dp[i - 1][j - 1][k]
            );
          } else {
            dp[i][j][k] = dp[i - 1][j - 1][k];
          }
        } else {
          // Case 2: current elements of both arrays are different
          // Choose the longest subsequence without the current element
          dp[i][j][k] = Math.max(dp[i - 1][j][k], dp[i][j - 1][k]);
        }
      }
    }
  }
  // Return the final answer, or -1 if it is impossible
  return dp[n][m][sum] < 0 ? -1 : dp[n][m][sum];
}

const a = [9, 11, 2, 1, 6, 2, 7];
const b = [1, 2, 6, 9, 2, 3, 11, 7];
const n = a.length;
const m = b.length;
const sum = 18;

// function call
const ans = solve(a, b, n, m, sum);
console.log(ans === -1 ? -1 : ans); // Print the answer or -1 if it is impossible

Output:

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

Longest Common Subsequence with no repeating character

prince_patel27

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Length of Longest Common Subsequence with given sum K

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