Check if there exist 4 indices in the array satisfying the given condition

Given an array A[] of N positive integers and 3 integers X, Y, and Z, the task is to check if there exist 4 indices (say p, q, r, s) such that the following conditions are satisfied:

• 0 < p < q < r < s < N
• Sum of the subarray from A[p] to A[q - 1] is X
• Sum of the subarray from A[q] to A[r - 1] is Y
• Sum of the subarray from A[r] to A[s - 1] is Z

Examples:

Input: N = 10, A[] = {1, 3, 2, 2, 2, 3, 1, 4, 3, 2}, X = 5, Y = 7, Z = 5
Output: YES
Explanation: The 4 integers p, q, r, s are {1, 3, 6, 8}. 

• A[1] + A[2] = 5
• A[3] + A [4] + A[5] = 7
• A[6] + A[7] = 5

Input:  N = 9, A[] = {31, 41, 59, 26, 53, 58, 97, 93, 23}, X = 100, Y = 101, Z = 100
Output: NO

Approach: The problem can be solved easily with the help of cumulative sum and Binary search.

If we calculate the cumulative sum of the array, then the sum of any subarray can be computed in O(1). Say S[i] is cumulative sum till ith index, then S[j] - S[i] = A[i] + A[i + 1] + .... + A[j - 1].

So, given conditions can be converted into the following:

We need to find 4 integers p, q, r, s such that: 

S[q] - S[p] = X
S[r] - S[q] = Y
S[s] - S[r] = Z

Now, for any fixed index (say p), we can find another index (say q) using binary search in a monotonically increasing array (cumulative sum), such that S[q] = S[p] + X. Similarly, we can find r and s. We can perform this search for all possible indices.

NOTE: A set can be used so that we won't need to perform a binary search explicitly each time.

Thus, the problem can be solved using the following steps :

• Declare a set (say S).
• Initialize a variable (say curr) by 0, to calculate the cumulative sum at each iteration.
• Iterate through the given array and insert the cumulative sum into the set.
• Iterate through the set and check if the current element of the set satisfies the given condition along with 3 other elements (which are also in the set). If so, return true.
• Else, return false.

Below is the implementation for the above approach:

// C++ code based on the above approach

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

// Function to check if it is possible to
// find 4 integers satisfying the
// given condition
bool isPossible(int N, int A[], int X, int Y, int Z)
{
    // Declaring a set
    set<int> S({ 0 });

    // Initializing variable to store
    // cumulative sum
    int curr = 0;

    // Inserting cumulative sums
    // into the set
    for (int i = 0; i < N; i++) {
        curr += A[i];
        S.insert(curr);
    }

    // Iterating through the set
    for (auto it : S) {

        // If current element of set
        // satisfies  the given condition
        // along with 3 other elements
        // (which are also in set),
        // return true
        if (S.find(it + X) != S.end()
            && S.find(it + X + Y) != S.end()
            && S.find(it + X + Y + Z) != S.end()) {
            return true;
        }
    }

    // Return false if the iteration
    // completes without getting
    // the required elements
    return false;
}

// Driver code
int main()
{
    int N = 10, X = 5, Y = 7, Z = 5;
    int A[] = { 1, 3, 2, 2, 2, 3, 1, 4, 3, 2 };

    // Function call
    int answer = isPossible(N, A, X, Y, Z);
    if (answer == true) {
        cout << "YES" << endl;
    }
    else {
        cout << "NO" << endl;
    }
    return 0;
}

// Java code based on the above approach
import java.util.*;

class GFG{

  // Function to check if it is possible to
  // find 4 integers satisfying the
  // given condition
  static boolean isPossible(int N, int A[], int X, int Y, int Z)
  {
    
    // Declaring a set
    HashSet<Integer> S = new HashSet<>();

    // Initializing variable to store
    // cumulative sum
    int curr = 0;

    // Inserting cumulative sums
    // into the set
    for (int i = 0; i < N; i++) {
      curr += A[i];
      S.add(curr);
    }

    // Iterating through the set
    for (int it : S) {

      // If current element of set
      // satisfies  the given condition
      // along with 3 other elements
      // (which are also in set),
      // return true
      if (!S.contains(it + X)
          && !S.contains(it + X + Y) 
          && !S.contains(it + X + Y + Z) ) {
        return true;
      }
    }

    // Return false if the iteration
    // completes without getting
    // the required elements
    return false;
  }

  // Driver code
  public static void main(String[] args)
  {
    int N = 10, X = 5, Y = 7, Z = 5;
    int A[] = { 1, 3, 2, 2, 2, 3, 1, 4, 3, 2 };

    // Function call
    boolean answer = isPossible(N, A, X, Y, Z);
    if (answer == true) {
      System.out.print("YES" +"\n");
    }
    else {
      System.out.print("NO" +"\n");
    }
  }
}

// This code is contributed by shikhasingrajput 

# Function to check if it is possible to
# find 4 integers satisfying the
# given condition
def isPossible(N, A, X, Y, Z) :
    
    # Declaring a set
    S = set()

    # Initializing variable to store
    # cumulative sum
    curr = 0

    # Inserting cumulative sums
    # into the set
    for i in range(N): 
        curr += A[i]
        S.add(curr)
    

    # Iterating through the set
    for it in S:

        # If current element of set
        # satisfies  the given condition
        # along with 3 other elements
        # (which are also in set),
        # return true
        if ((it + X) in S
            and (it + X + Y) in S
            and (it + X + Y + Z) in S) :
            return True
        
    # Return false if the iteration
    # completes without getting
    # the required elements
    return False

# Driver code
if __name__ == "__main__":
    
    N = 10
    X = 5
    Y = 7
    Z = 5
    A = [ 1, 3, 2, 2, 2, 3, 1, 4, 3, 2 ]

    # Function call
    answer = isPossible(N, A, X, Y, Z)
    if (answer == True) :
        print("YES" )
    
    else :
        print("NO")

# This code is contributed by code_hunt.

// C# program for the above approach
using System;
using System.Collections.Generic;

public class GFG{

  // Function to check if it is possible to
  // find 4 integers satisfying the
  // given condition
  static bool isPossible(int N, int[] A, int X, int Y, int Z)
  {
    
    // Declaring a set
    HashSet<int> S = new HashSet<int>();

    // Initializing variable to store
    // cumulative sum
    int curr = 0;

    // Inserting cumulative sums
    // into the set
    for (int i = 0; i < N; i++) {
      curr += A[i];
      S.Add(curr);
    }

    // Iterating through the set
    foreach (int it in S) {

      // If current element of set
      // satisfies  the given condition
      // along with 3 other elements
      // (which are also in set),
      // return true
      if (!S.Contains(it + X)
          && !S.Contains(it + X + Y) 
          && !S.Contains(it + X + Y + Z) ) {
        return true;
      }
    }

    // Return false if the iteration
    // completes without getting
    // the required elements
    return false;
  }

static public void Main ()
{
    int N = 10, X = 5, Y = 7, Z = 5;
    int[] A = { 1, 3, 2, 2, 2, 3, 1, 4, 3, 2 };

    // Function call
    bool answer = isPossible(N, A, X, Y, Z);
    if (answer == true) {
      Console.Write("YES" +"\n");
    }
    else {
      Console.Write("NO" +"\n");
    }
}
}

// This code is contributed by sanjoy_62.

// JavaScript code based on the above approach

// Function to check if it is possible to
// find 4 letters satisfying the
// given condition
function isPossible(N, A, X, Y, Z) 
{
  // Declaring a set
  let S = new Set([0]);

  // Initializing variable to store
  // cumulative sum
  let curr = 0;

  // adding cumulative sums
  // into the set
  for (let i = 0; i < N; i++) {
    curr += A[i];
    S.add(curr);
  }

  // Iterating through the set
  for (const it of S.values()) {
    // If current element of set
    // satisfies  the given condition
    // along with 3 other elements
    // (which are also in set),
    // return true
    if (S.has(it + X) && S.has(it + X + Y) && S.has(it + X + Y + Z)) {
      return true;
    }
  }

  // Return false if the iteration
  // completes without getting
  // the required elements
  return false;
}

// Driver code
let N = 10,
  X = 5,
  Y = 7,
  Z = 5;
let A = [1, 3, 2, 2, 2, 3, 1, 4, 3, 2];

// Function call
let answer = isPossible(N, A, X, Y, Z);
if (answer == true) {
  console.log("YES");
} else {
  console.log("NO");
}

// This code is contributed by ishankhandelwals.

YES

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

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Article Tags :

• Searching
• DSA
• Arrays
• Binary Search
• prefix-sum
• cpp-set

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Practice Tags :

• Arrays
• Binary Search
• prefix-sum
• Searching