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# Print all the combinations of N elements by changing sign such that their sum is divisible by M

• Difficulty Level : Hard
• Last Updated : 29 Aug, 2022

Given an array of N integers and an integer M. You can change the sign(positive or negative) of any element in the array. The task is to print all possible combinations of the array elements that can be obtained by changing the sign of the elements such that their sum is divisible by M.

Note: You have to take all of the array elements in each combination and in the same order as the elements present in the array. However, you can change the sign of elements.

Examples:

Input: a[] = {5, 6, 7}, M = 3
Output:
-5-6-7
+5-6+7
-5+6-7
+5+6+7

Input: a[] = {3, 5, 6, 8}, M = 5
Output:
-3-5+6-8
-3+5+6-8
+3-5-6+8
+3+5-6+8

Approach: The concept of power set is used here to solve this problem. Using power-set generate all possible combinations of signs that can be applied to the array of elements. If the sum obtained is divisible by M, then print the combination. Below are the steps:

• Iterate for all possible combinations of ‘+’ and ‘-‘ using power set.
• Iterate on the array elements and if the j-th bit from left is set, then assume the array element to be positive and if the bit is not set, then assume the array element to be negative. Refer here to check if bit any index is set or not.
• If the sum is divisible by M, then the again traverse the array elements and print them along with sign( ‘+’ or ‘-‘ ).

Below is the implementation of the above approach:

## C++

 #include using namespace std;   // Function to print all the combinations void printCombinations(int a[], int n, int m) {       // Iterate for all combinations     for (int i = 0; i < (1 << n); i++) {         int sum = 0;           // Initially 100 in binary if n is 3         // as 1<<(3-1) = 100 in binary         int num = 1 << (n - 1);           // Iterate in the array and assign signs         // to the array elements         for (int j = 0; j < n; j++) {               // If the j-th bit from left is set             // take '+' sign             if (i & num)                 sum += a[j];             else                 sum += (-1 * a[j]);               // Right shift to check if             // jth bit is set or not             num = num >> 1;         }           if (sum % m == 0) {               // re-initialize             num = 1 << (n - 1);               // Iterate in the array elements             for (int j = 0; j < n; j++) {                   // If the jth from left is set                 if ((i & num))                     cout << "+" << a[j] << " ";                 else                     cout << "-" << a[j] << " ";                   // right shift                 num = num >> 1;             }             cout << endl;         }     } }   // Driver Code int main() {     int a[] = { 3, 5, 6, 8 };     int n = sizeof(a) / sizeof(a[0]);     int m = 5;       printCombinations(a, n, m);     return 0; }

## Java

 import java.io.*;   class GFG {       // Function to print // all the combinations static void printCombinations(int a[],                               int n, int m) {       // Iterate for all     // combinations     for (int i = 0;              i < (1 << n); i++)     {         int sum = 0;           // Initially 100 in binary         // if n is 3 as         // 1<<(3-1) = 100 in binary         int num = 1 << (n - 1);           // Iterate in the array         // and assign signs to         // the array elements         for (int j = 0; j < n; j++)         {               // If the j-th bit             // from left is set             // take '+' sign             if ((i & num) > 0)                 sum += a[j];             else                 sum += (-1 * a[j]);               // Right shift to check if             // jth bit is set or not             num = num >> 1;         }           if (sum % m == 0)         {               // re-initialize             num = 1 << (n - 1);               // Iterate in the             // array elements             for (int j = 0; j < n; j++)             {                   // If the jth from                 // left is set                 if ((i & num) > 0)                     System.out.print("+" +                                      a[j] + " ");                 else                     System.out.print("-" +                                      a[j] + " ");                   // right shift                 num = num >> 1;             }                       System.out.println();         }     } }   // Driver code public static void main(String args[]) {     int a[] = { 3, 5, 6, 8 };     int n = a.length;     int m = 5;       printCombinations(a, n, m); } }   // This code is contributed // by inder_verma.

## Python3

 # Function to print # all the combinations def printCombinations(a, n, m):       # Iterate for all     # combinations     for i in range(0, (1 << n)):               sum = 0           # Initially 100 in binary         # if n is 3 as         # 1<<(3-1) = 100 in binary         num = 1 << (n - 1)           # Iterate in the array         # and assign signs to         # the array elements         for j in range(0, n):                       # If the j-th bit             # from left is set             # take '+' sign             if ((i & num) > 0):                 sum += a[j]             else:                 sum += (-1 * a[j])               # Right shift to check if             # jth bit is set or not             num = num >> 1           if (sum % m == 0):                       # re-initialize             num = 1 << (n - 1)               # Iterate in the             # array elements             for j in range(0, n):                   # If the jth from                 # left is set                 if ((i & num) > 0):                     print("+", a[j], end = " ",                                      sep = "")                 else:                     print("-", a[j], end = " ",                                      sep = "")                   # right shift                 num = num >> 1             print("")       # Driver code a = [ 3, 5, 6, 8 ] n = len(a) m = 5 printCombinations(a, n, m)   # This code is contributed # by smita.

## C#

 // Print all the combinations // of N elements by changing // sign such that their sum // is divisible by M using System;   class GFG {       // Function to print // all the combinations static void printCombinations(int []a,                               int n, int m) {       // Iterate for all     // combinations     for (int i = 0;              i < (1 << n); i++)     {         int sum = 0;           // Initially 100 in binary         // if n is 3 as         // 1<<(3-1) = 100 in binary         int num = 1 << (n - 1);           // Iterate in the array         // and assign signs to         // the array elements         for (int j = 0; j < n; j++)         {               // If the j-th bit             // from left is set             // take '+' sign             if ((i & num) > 0)                 sum += a[j];             else                 sum += (-1 * a[j]);               // Right shift to check if             // jth bit is set or not             num = num >> 1;         }           if (sum % m == 0)         {               // re-initialize             num = 1 << (n - 1);               // Iterate in the             // array elements             for (int j = 0; j < n; j++)             {                   // If the jth from                 // left is set                 if ((i & num) > 0)                     Console.Write("+" +                               a[j] + " ");                 else                     Console.Write("-" +                               a[j] + " ");                   // right shift                 num = num >> 1;             }                       Console.Write("\n");         }     } }   // Driver code public static void Main() {     int []a = { 3, 5, 6, 8 };     int n = a.Length;     int m = 5;       printCombinations(a, n, m); } }   // This code is contributed // by Smitha.

## PHP

 > 1;         }           if (\$sum % \$m == 0)         {               // re-initialize             \$num = 1 << (\$n - 1);               // Iterate in the array elements             for (\$j = 0; \$j < \$n; \$j++)             {                   // If the jth from left is set                 if ((\$i & \$num))                     echo "+" , \$a[\$j] , " ";                 else                     echo "-" , \$a[\$j] , " ";                   // right shift                 \$num = \$num >> 1;             }         echo "\n";         }     } }   // Driver Code \$a = array( 3, 5, 6, 8 ); \$n = sizeof(\$a); \$m = 5;   printCombinations(\$a, \$n, \$m);   // This code is contributed by ajit ?>

## Javascript



Output

-3 -5 +6 -8
-3 +5 +6 -8
+3 -5 -6 +8
+3 +5 -6 +8

Time Complexity: O(2N * N), where N is the number of elements.

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