Given an array arr[] of n numbers and a number K, find the number of subsets of arr[] having XOR of elements as K**Examples :**

Input: arr[] = {6, 9, 4,2}, k = 6 Output: 2 The subsets are {4, 2} and {6} Input: arr[] = {1, 2, 3, 4, 5}, k = 4 Output: 4 The subsets are {1, 5}, {4}, {1, 2, 3, 4} and {2, 3, 5}

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**Brute Force approach O(2 ^{n}): **One naive approach is to generate all the 2

^{n}subsets and count all the subsets having XOR value K, but this approach will not be efficient for large values of n.

**Dynamic Programming Approach O(n*m): **

We define a number m such that m = pow(2,(log2(max(arr))+1)) – 1. This number is actually the maximum value any XOR subset will acquire. We get this number by counting bits in largest number. We create a 2D array dp[n+1][m+1], such that** dp[i][j] equals to the number of subsets having XOR value j from subsets of arr[0…i-1]**.

We fill the dp array as following:

- We initialize all values of dp[i][j] as 0.
- Set value of dp[0][0] = 1 since XOR of an empty set is 0.
- Iterate over all the values of arr[i] from left to right and for each arr[i], iterate over all the possible values of XOR i.e from 0 to m (both inclusive) and fill the dp array asfollowing:

for i = 1 to n:

for j = 0 to m:

dp[i][j] = dp[i-1][j] + dp[i-1][j^arr[i-1]]

This can be explained as, if there is a subset arr[0…i-2] with XOR value j, then there also exists a subset arr[0…i-1] with XOR value j. also if there exists a subset arr[0….i-2] with XOR value j^arr[i] then clearly there exist a subset arr[0…i-1] with XOR value j, as j ^ arr[i-1] ^ arr[i-1] = j. - Counting the number of subsets with XOR value k: Since dp[i][j] is the number of subsets having j as XOR value from the subsets of arr[0..i-1], then the number of subsets from set arr[0..n] having XOR value as K will be dp[n][K]

## C++

`// arr dynamic programming solution to finding the number` `// of subsets having xor of their elements as k` `#include<bits/stdc++.h>` `using` `namespace` `std;` `// Returns count of subsets of arr[] with XOR value equals` `// to k.` `int` `subsetXOR(` `int` `arr[], ` `int` `n, ` `int` `k)` `{` ` ` `// Find maximum element in arr[]` ` ` `int` `max_ele = arr[0];` ` ` `for` `(` `int` `i=1; i<n; i++)` ` ` `if` `(arr[i] > max_ele)` ` ` `max_ele = arr[i];` ` ` `// Maximum possible XOR value` ` ` `int` `m = (1 << (` `int` `)(log2(max_ele) + 1) ) - 1;` ` ` `if` `( k > m )` ` ` `return` `0;` ` ` `// The value of dp[i][j] is the number of subsets having` ` ` `// XOR of their elements as j from the set arr[0...i-1]` ` ` `int` `dp[n+1][m+1];` ` ` ` ` `// Initializing all the values of dp[i][j] as 0` ` ` `for` `(` `int` `i=0; i<=n; i++)` ` ` `for` `(` `int` `j=0; j<=m; j++)` ` ` `dp[i][j] = 0;` ` ` `// The xor of empty subset is 0` ` ` `dp[0][0] = 1;` ` ` `// Fill the dp table` ` ` `for` `(` `int` `i=1; i<=n; i++)` ` ` `for` `(` `int` `j=0; j<=m; j++)` ` ` `dp[i][j] = dp[i-1][j] + dp[i-1][j^arr[i-1]];` ` ` `// The answer is the number of subset from set` ` ` `// arr[0..n-1] having XOR of elements as k` ` ` `return` `dp[n][k];` `}` `// Driver program to test above function` `int` `main()` `{` ` ` `int` `arr[] = {1, 2, 3, 4, 5};` ` ` `int` `k = 4;` ` ` `int` `n = ` `sizeof` `(arr)/` `sizeof` `(arr[0]);` ` ` `cout << ` `"Count of subsets is "` `<< subsetXOR(arr, n, k);` ` ` `return` `0;` `}` |

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## Java

`// Java dynamic programming solution` `// to finding the number of subsets ` `// having xor of their elements as k ` `class` `GFG{ ` ` ` `// Returns count of subsets of arr[] with ` `// XOR value equals to k. ` `static` `int` `subsetXOR(` `int` `[]arr, ` `int` `n, ` `int` `k) ` `{ ` ` ` ` ` `// Find maximum element in arr[] ` ` ` `int` `max_ele = arr[` `0` `]; ` ` ` ` ` `for` `(` `int` `i = ` `1` `; i < n; i++) ` ` ` `if` `(arr[i] > max_ele) ` ` ` `max_ele = arr[i]; ` ` ` ` ` `// Maximum possible XOR value ` ` ` `int` `m = (` `1` `<< (` `int` `)(Math.log(max_ele) / ` ` ` `Math.log(` `2` `) + ` `1` `) ) - ` `1` `; ` ` ` `if` `(k > m)` ` ` `{ ` ` ` `return` `0` `; ` ` ` `} ` ` ` ` ` `// The value of dp[i][j] is the number` ` ` `// of subsets having XOR of their ` ` ` `// elements as j from the set arr[0...i-1] ` ` ` `int` `[][]dp = ` `new` `int` `[n + ` `1` `][m + ` `1` `]; ` ` ` ` ` `// Initializing all the values of dp[i][j] as 0 ` ` ` `for` `(` `int` `i = ` `0` `; i <= n; i++) ` ` ` `for` `(` `int` `j = ` `0` `; j <= m; j++) ` ` ` `dp[i][j] = ` `0` `; ` ` ` ` ` `// The xor of empty subset is 0 ` ` ` `dp[` `0` `][` `0` `] = ` `1` `; ` ` ` ` ` `// Fill the dp table ` ` ` `for` `(` `int` `i = ` `1` `; i <= n; i++) ` ` ` `for` `(` `int` `j = ` `0` `; j <= m; j++) ` ` ` `dp[i][j] = dp[i - ` `1` `][j] + ` ` ` `dp[i - ` `1` `][j ^ arr[i - ` `1` `]]; ` ` ` ` ` `// The answer is the number of ` ` ` `// subset from set arr[0..n-1]` ` ` `// having XOR of elements as k ` ` ` `return` `dp[n][k]; ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String arg[]) ` `{ ` ` ` `int` `[]arr = { ` `1` `, ` `2` `, ` `3` `, ` `4` `, ` `5` `}; ` ` ` `int` `k = ` `4` `; ` ` ` `int` `n = arr.length; ` ` ` ` ` `System.out.println(` `"Count of subsets is "` `+` ` ` `subsetXOR(arr, n, k)); ` `} ` `} ` `// This code is contributed by rutvik_56` |

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## Python3

`# Python 3 arr dynamic programming solution ` `# to finding the number of subsets having ` `# xor of their elements as k` `import` `math` `# Returns count of subsets of arr[] with ` `# XOR value equals to k.` `def` `subsetXOR(arr, n, k):` ` ` ` ` `# Find maximum element in arr[]` ` ` `max_ele ` `=` `arr[` `0` `]` ` ` `for` `i ` `in` `range` `(` `1` `, n):` ` ` `if` `arr[i] > max_ele :` ` ` `max_ele ` `=` `arr[i]` ` ` ` ` `# Maximum possible XOR value` ` ` `m ` `=` `(` `1` `<< (` `int` `)(math.log2(max_ele) ` `+` `1` `)) ` `-` `1` ` ` `if` `( k > m ):` ` ` `return` `0` ` ` `# The value of dp[i][j] is the number ` ` ` `# of subsets having XOR of their elements ` ` ` `# as j from the set arr[0...i-1]` ` ` `# Initializing all the values ` ` ` `# of dp[i][j] as 0` ` ` `dp ` `=` `[[` `0` `for` `i ` `in` `range` `(m ` `+` `1` `)]` ` ` `for` `i ` `in` `range` `(n ` `+` `1` `)]` ` ` ` ` `# The xor of empty subset is 0` ` ` `dp[` `0` `][` `0` `] ` `=` `1` ` ` `# Fill the dp table` ` ` `for` `i ` `in` `range` `(` `1` `, n ` `+` `1` `):` ` ` `for` `j ` `in` `range` `(m ` `+` `1` `):` ` ` `dp[i][j] ` `=` `(dp[i ` `-` `1` `][j] ` `+` ` ` `dp[i ` `-` `1` `][j ^ arr[i ` `-` `1` `]])` ` ` `# The answer is the number of subset ` ` ` `# from set arr[0..n-1] having XOR of` ` ` `# elements as k` ` ` `return` `dp[n][k]` `# Driver Code` `arr ` `=` `[` `1` `, ` `2` `, ` `3` `, ` `4` `, ` `5` `]` `k ` `=` `4` `n ` `=` `len` `(arr)` `print` `(` `"Count of subsets is"` `, ` ` ` `subsetXOR(arr, n, k))` `# This code is contributed` `# by sahishelangia` |

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## C#

`// C# dynamic programming solution to finding the number` `// of subsets having xor of their elements as k` `using` `System;` `class` `GFG` `{` ` ` `// Returns count of subsets of arr[] with ` `// XOR value equals to k.` `static` `int` `subsetXOR(` `int` `[]arr, ` `int` `n, ` `int` `k)` `{` ` ` `// Find maximum element in arr[]` ` ` `int` `max_ele = arr[0];` ` ` `for` `(` `int` `i = 1; i < n; i++)` ` ` `if` `(arr[i] > max_ele)` ` ` `max_ele = arr[i];` ` ` `// Maximum possible XOR value` ` ` `int` `m = (1 << (` `int` `)(Math.Log(max_ele,2) + 1) ) - 1;` ` ` `if` `( k > m ){` ` ` `return` `0; ` ` ` `}` ` ` `// The value of dp[i][j] is the number of subsets having` ` ` `// XOR of their elements as j from the set arr[0...i-1]` ` ` `int` `[,]dp=` `new` `int` `[n+1,m+1];` ` ` `// Initializing all the values of dp[i][j] as 0` ` ` `for` `(` `int` `i = 0; i <= n; i++)` ` ` `for` `(` `int` `j = 0; j <= m; j++)` ` ` `dp[i, j] = 0;` ` ` `// The xor of empty subset is 0` ` ` `dp[0, 0] = 1;` ` ` `// Fill the dp table` ` ` `for` `(` `int` `i = 1; i <= n; i++)` ` ` `for` `(` `int` `j = 0; j <= m; j++)` ` ` `dp[i, j] = dp[i-1, j] + dp[i-1, j^arr[i-1]];` ` ` `// The answer is the number of subset from set` ` ` `// arr[0..n-1] having XOR of elements as k` ` ` `return` `dp[n, k];` `}` ` ` `// Driver code` ` ` `static` `public` `void` `Main ()` ` ` `{` ` ` `int` `[]arr = {1, 2, 3, 4, 5};` ` ` `int` `k = 4;` ` ` `int` `n = arr.Length;` ` ` `Console.WriteLine (` `"Count of subsets is "` `+ subsetXOR(arr, n, k));` ` ` `}` `}` `// This code is contributed by jit_t.` |

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## PHP

`<?php` `// PHP arr dynamic programming ` `// solution to finding the number ` `// of subsets having xor of their ` `// elements as k` `// Returns count of subsets of ` `// arr[] with XOR value equals to k.` `function` `subsetXOR(` `$arr` `, ` `$n` `, ` `$k` `)` `{` ` ` `// Find maximum element in arr[]` ` ` `$max_ele` `= ` `$arr` `[0];` ` ` `for` `(` `$i` `= 1; ` `$i` `< ` `$n` `; ` `$i` `++)` ` ` `if` `(` `$arr` `[` `$i` `] > ` `$max_ele` `)` ` ` `$max_ele` `= ` `$arr` `[` `$i` `];` ` ` `// Maximum possible XOR value` ` ` `$m` `= (1 << (int)(log(` `$max_ele` `, ` ` ` `2) + 1) ) - 1;` ` ` `if` `( ` `$k` `> ` `$m` `){` ` ` `return` `0;` ` ` `}` ` ` `// The value of dp[i][j] is the` ` ` `// number of subsets having` ` ` `// XOR of their elements as j ` ` ` `// from the set arr[0...i-1]` ` ` ` ` `// Initializing all the ` ` ` `// values of dp[i][j] as 0` ` ` `for` `(` `$i` `= 0; ` `$i` `<= ` `$n` `; ` `$i` `++)` ` ` `for` `(` `$j` `= 0; ` `$j` `<= ` `$m` `; ` `$j` `++)` ` ` `$dp` `[` `$i` `][` `$j` `] = 0;` ` ` `// The xor of empty subset is 0` ` ` `$dp` `[0][0] = 1;` ` ` `// Fill the dp table` ` ` `for` `(` `$i` `= 1; ` `$i` `<= ` `$n` `; ` `$i` `++)` ` ` `for` `( ` `$j` `= 0; ` `$j` `<= ` `$m` `; ` `$j` `++)` ` ` `$dp` `[` `$i` `][` `$j` `] = ` `$dp` `[` `$i` `- 1][` `$j` `] + ` ` ` `$dp` `[` `$i` `- 1][` `$j` `^ ` ` ` `$arr` `[` `$i` `- 1]];` ` ` `// The answer is the number ` ` ` `// of subset from set arr[0..n-1] ` ` ` `// having XOR of elements as k` ` ` `return` `$dp` `[` `$n` `][` `$k` `];` `}` `// Driver Code` `$arr` `= ` `array` `(1, 2, 3, 4, 5);` `$k` `= 4;` `$n` `= sizeof(` `$arr` `);` `echo` `"Count of subsets is "` `, ` ` ` `subsetXOR(` `$arr` `, ` `$n` `, ` `$k` `);` `// This code is contributed by ajit` `?>` |

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**Output :**

Count of subsets is 4

This article is contributed by **Pranay Pandey**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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