Find four elements that sum to a given value | Set 1 (n^3 solution)
Given an array of integers, find all combination of four elements in the array whose sum is equal to a given value X.
For example, if the given array is {10, 2, 3, 4, 5, 9, 7, 8} and X = 23, then your function should print “3 5 7 8” (3 + 5 + 7 + 8 = 23).
A Naive Solution is to generate all possible quadruples and compare the sum of every quadruple with X. The following code implements this simple method using four nested loops
C++
// C++ program for naive solution to // print all combination of 4 elements // in A[] with sum equal to X #include <bits/stdc++.h> using namespace std; /* A naive solution to print all combination of 4 elements in A[]with sum equal to X */void findFourElements(int A[], int n, int X) { // Fix the first element and find other three for (int i = 0; i < n - 3; i++) { // Fix the second element and find other two for (int j = i + 1; j < n - 2; j++) { // Fix the third element and find the fourth for (int k = j + 1; k < n - 1; k++) { // find the fourth for (int l = k + 1; l < n; l++) if (A[i] + A[j] + A[k] + A[l] == X) cout << A[i] <<", " << A[j] << ", " << A[k] << ", " << A[l]; } } } } // Driver Code int main() { int A[] = {10, 20, 30, 40, 1, 2}; int n = sizeof(A) / sizeof(A[0]); int X = 91; findFourElements (A, n, X); return 0; } // This code is contributed // by Akanksha Rai |
C
#include <stdio.h> /* A naive solution to print all combination of 4 elements in A[] with sum equal to X */void findFourElements(int A[], int n, int X) { // Fix the first element and find other three for (int i = 0; i < n-3; i++) { // Fix the second element and find other two for (int j = i+1; j < n-2; j++) { // Fix the third element and find the fourth for (int k = j+1; k < n-1; k++) { // find the fourth for (int l = k+1; l < n; l++) if (A[i] + A[j] + A[k] + A[l] == X) printf("%d, %d, %d, %d", A[i], A[j], A[k], A[l]); } } } } // Driver program to test above function int main() { int A[] = {10, 20, 30, 40, 1, 2}; int n = sizeof(A) / sizeof(A[0]); int X = 91; findFourElements (A, n, X); return 0; } |
Java
class FindFourElements { /* A naive solution to print all combination of 4 elements in A[] with sum equal to X */ void findFourElements(int A[], int n, int X) { // Fix the first element and find other three for (int i = 0; i < n - 3; i++) { // Fix the second element and find other two for (int j = i + 1; j < n - 2; j++) { // Fix the third element and find the fourth for (int k = j + 1; k < n - 1; k++) { // find the fourth for (int l = k + 1; l < n; l++) { if (A[i] + A[j] + A[k] + A[l] == X) System.out.print(A[i]+" "+A[j]+" "+A[k] +" "+A[l]); } } } } } // Driver program to test above functions public static void main(String[] args) { FindFourElements findfour = new FindFourElements(); int A[] = {10, 20, 30, 40, 1, 2}; int n = A.length; int X = 91; findfour.findFourElements(A, n, X); } } |
Python3
# A naive solution to print all combination # of 4 elements in A[] with sum equal to X def findFourElements(A, n, X): # Fix the first element and find # other three for i in range(0,n-3): # Fix the second element and # find other two for j in range(i+1,n-2): # Fix the third element # and find the fourth for k in range(j+1,n-1): # find the fourth for l in range(k+1,n): if A[i] + A[j] + A[k] + A[l] == X: print ("%d, %d, %d, %d" %( A[i], A[j], A[k], A[l])) # Driver program to test above function A = [10, 2, 3, 4, 5, 9, 7, 8] n = len(A) X = 23findFourElements (A, n, X) # This code is contributed by shreyanshi_arun |
C#
// C# program for naive solution to // print all combination of 4 elements // in A[] with sum equal to X using System; class FindFourElements { void findFourElements(int []A, int n, int X) { // Fix the first element and find other three for (int i = 0; i < n - 3; i++) { // Fix the second element and find other two for (int j = i + 1; j < n - 2; j++) { // Fix the third element and find the fourth for (int k = j + 1; k < n - 1; k++) { // find the fourth for (int l = k + 1; l < n; l++) { if (A[i] + A[j] + A[k] + A[l] == X) Console.Write(A[i] + " " + A[j] + " " + A[k] + " " + A[l]); } } } } } // Driver program to test above functions public static void Main() { FindFourElements findfour = new FindFourElements(); int []A = {10, 20, 30, 40, 1, 2}; int n = A.Length; int X = 91; findfour.findFourElements(A, n, X); } } // This code is contributed by nitin mittal |
PHP
<?php /* A naive solution to print all combination of 4 elements in A[] with sum equal to X */ function findFourElements($A, $n, $X) { // Fix the first element and find other // three for ($i = 0; $i < $n-3; $i++) { // Fix the second element and find // other two for ($j = $i+1; $j < $n-2; $j++) { // Fix the third element and // find the fourth for ($k = $j+1; $k < $n-1; $k++) { // find the fourth for ($l = $k+1; $l < $n; $l++) if ($A[$i] + $A[$j] + $A[$k] + $A[$l] == $X) echo $A[$i], ", " , $A[$j], ", ", $A[$k], ", ", $A[$l]; } } } } // Driver program to test above function $A = array (10, 20, 30, 40, 1, 2); $n = sizeof($A) ; $X = 91; findFourElements ($A, $n, $X); // This code is contributed by m_kit ?> |
Output:
20, 30, 40, 1
Time Complexity: O(n^4)
The time complexity can be improved to O(n^3) with the use of sorting as a preprocessing step, and then using method 1 of this post to reduce a loop.
Following are the detailed steps.
1) Sort the input array.
2) Fix the first element as A[i] where i is from 0 to n–3. After fixing the first element of quadruple, fix the second element as A[j] where j varies from i+1 to n-2. Find remaining two elements in O(n) time, using the method 1 of this post
Following is the implementation of O(n^3) solution.
C++
// C++ program for to print all combination // of 4 elements in A[] with sum equal to X #include<bits/stdc++.h> using namespace std; /* Following function is needed for library function qsort(). */int compare (const void *a, const void * b) { return ( *(int *)a - *(int *)b ); } /* A sorting based solution to print all combination of 4 elements in A[] with sum equal to X */void find4Numbers(int A[], int n, int X) { int l, r; // Sort the array in increasing // order, using library function // for quick sort qsort (A, n, sizeof(A[0]), compare); /* Now fix the first 2 elements one by one and find the other two elements */ for (int i = 0; i < n - 3; i++) { for (int j = i+1; j < n - 2; j++) { // Initialize two variables as // indexes of the first and last // elements in the remaining elements l = j + 1; r = n-1; // To find the remaining two // elements, move the index // variables (l & r) toward each other. while (l < r) { if( A[i] + A[j] + A[l] + A[r] == X) { cout << A[i]<<", " << A[j] << ", " << A[l] << ", " << A[r]; l++; r--; } else if (A[i] + A[j] + A[l] + A[r] < X) l++; else // A[i] + A[j] + A[l] + A[r] > X r--; } // end of while } // end of inner for loop } // end of outer for loop } /* Driver code */int main() { int A[] = {1, 4, 45, 6, 10, 12}; int X = 21; int n = sizeof(A) / sizeof(A[0]); find4Numbers(A, n, X); return 0; } // This code is contributed by rathbhupendra |
C
# include <stdio.h> # include <stdlib.h> /* Following function is needed for library function qsort(). Refer int compare (const void *a, const void * b) { return ( *(int *)a - *(int *)b ); } /* A sorting based solution to print all combination of 4 elements in A[] with sum equal to X */void find4Numbers(int A[], int n, int X) { int l, r; // Sort the array in increasing order, using library // function for quick sort qsort (A, n, sizeof(A[0]), compare); /* Now fix the first 2 elements one by one and find the other two elements */ for (int i = 0; i < n - 3; i++) { for (int j = i+1; j < n - 2; j++) { // Initialize two variables as indexes of the first and last // elements in the remaining elements l = j + 1; r = n-1; // To find the remaining two elements, move the index // variables (l & r) toward each other. while (l < r) { if( A[i] + A[j] + A[l] + A[r] == X) { printf("%d, %d, %d, %d", A[i], A[j], A[l], A[r]); l++; r--; } else if (A[i] + A[j] + A[l] + A[r] < X) l++; else // A[i] + A[j] + A[l] + A[r] > X r--; } // end of while } // end of inner for loop } // end of outer for loop } /* Driver program to test above function */int main() { int A[] = {1, 4, 45, 6, 10, 12}; int X = 21; int n = sizeof(A)/sizeof(A[0]); find4Numbers(A, n, X); return 0; } |
Java
import java.util.Arrays; class FindFourElements { /* A sorting based solution to print all combination of 4 elements in A[] with sum equal to X */ void find4Numbers(int A[], int n, int X) { int l, r; // Sort the array in increasing order, using library // function for quick sort Arrays.sort(A); /* Now fix the first 2 elements one by one and find the other two elements */ for (int i = 0; i < n - 3; i++) { for (int j = i + 1; j < n - 2; j++) { // Initialize two variables as indexes of the first and last // elements in the remaining elements l = j + 1; r = n - 1; // To find the remaining two elements, move the index // variables (l & r) toward each other. while (l < r) { if (A[i] + A[j] + A[l] + A[r] == X) { System.out.println(A[i]+" "+A[j]+" "+A[l]+" "+A[r]); l++; r--; } else if (A[i] + A[j] + A[l] + A[r] < X) l++; else // A[i] + A[j] + A[l] + A[r] > X r--; } // end of while } // end of inner for loop } // end of outer for loop } // Driver program to test above functions public static void main(String[] args) { FindFourElements findfour = new FindFourElements(); int A[] = {1, 4, 45, 6, 10, 12}; int n = A.length; int X = 21; findfour.find4Numbers(A, n, X); } } // This code has been contributed by Mayank Jaiswal |
Python 3
# A sorting based solution to print all combination # of 4 elements in A[] with sum equal to X def find4Numbers(A, n, X): # Sort the array in increasing order, # using library function for quick sort A.sort() # Now fix the first 2 elements one by # one and find the other two elements for i in range(n - 3): for j in range(i + 1, n - 2): # Initialize two variables as indexes # of the first and last elements in # the remaining elements l = j + 1 r = n - 1 # To find the remaining two elements, # move the index variables (l & r) # toward each other. while (l < r): if(A[i] + A[j] + A[l] + A[r] == X): print(A[i], ",", A[j], ",", A[l], ",", A[r]) l += 1 r -= 1 elif (A[i] + A[j] + A[l] + A[r] < X): l += 1 else: # A[i] + A[j] + A[l] + A[r] > X r -= 1 # Driver Code if __name__ == "__main__": A = [1, 4, 45, 6, 10, 12] X = 21 n = len(A) find4Numbers(A, n, X) # This code is contributed by ita_c |
C#
// C# program for to print all combination // of 4 elements in A[] with sum equal to X using System; class FindFourElements { // A sorting based solution to print all // combination of 4 elements in A[] with // sum equal to X void find4Numbers(int []A, int n, int X) { int l, r; // Sort the array in increasing order, // using library function for quick sort Array.Sort(A); /* Now fix the first 2 elements one by one and find the other two elements */ for (int i = 0; i < n - 3; i++) { for (int j = i + 1; j < n - 2; j++) { // Initialize two variables as indexes of // the first and last elements in the // remaining elements l = j + 1; r = n - 1; // To find the remaining two elements, move the // index variables (l & r) toward each other. while (l < r) { if (A[i] + A[j] + A[l] + A[r] == X) { Console.Write(A[i] + " " + A[j] + " " + A[l] + " " + A[r]); l++; r--; } else if (A[i] + A[j] + A[l] + A[r] < X) l++; else // A[i] + A[j] + A[l] + A[r] > X r--; } // end of while } // end of inner for loop } // end of outer for loop } // Driver program to test above functions public static void Main() { FindFourElements findfour = new FindFourElements(); int []A = {1, 4, 45, 6, 10, 12}; int n = A.Length; int X = 21; findfour.find4Numbers(A, n, X); } } // This code has been contributed by nitin mittal |
PHP
<?php // PHP program for to print all // combination of 4 elements in // A[] with sum equal to X // A sorting based solution to // print all combination of 4 // elements in A[] with sum // equal to X function find4Numbers($A, $n, $X) { $l; $r; // Sort the array in increasing // order, using library function // for quick sort sort($A); // Now fix the first 2 elements // one by one and find the other // two elements for ($i = 0; $i < $n - 3; $i++) { for ($j = $i + 1; $j < $n - 2; $j++) { // Initialize two variables // as indexes of the first // and last elements in the // remaining elements $l = $j + 1; $r = $n - 1; // To find the remaining two // elements, move the index // variables (l & r) towards // each other. while ($l < $r) { if ($A[$i] + $A[$j] + $A[$l] + $A[$r] == $X) { echo $A[$i] , " " , $A[$j] , " " , $A[$l] , " " , $A[$r]; $l++; $r--; } else if ($A[$i] + $A[$j] + $A[$l] + $A[$r] < $X) $l++; // A[i] + A[j] + A[l] + A[r] > X else $r--; } } } } // Driver Code $A = array(1, 4, 45, 6, 10, 12); $n = count($A); $X = 21; find4Numbers($A, $n, $X); // This code is contributed // by nitin mittal ?> |
Output :
1, 4, 6, 10
Time Complexity : O(n^3)
This problem can also be solved in O(n^2Logn) complexity. Please refer below post for details
Find four elements that sum to a given value | Set 2 ( O(n^2Logn) Solution)
Please write comments if you find any of the above codes/algorithms incorrect, or find other ways to solve the same problem.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
