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 = 23 findFourElements (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.
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