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XORingSubarrays.java
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XORingSubarrays.java
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/*
Problem Description
Given an integer array A of size N.
You need to find the value obtained by XOR-ing the contiguous subarrays, followed by XOR-ing the values thus obtained. Determine and return this value.
For example, if A = [3, 4, 5] :
Subarray Operation Result
3 None 3
4 None 4
5 None 5
3,4 3 XOR 4 7
4,5 4 XOR 5 1
3,4,5 3 XOR 4 XOR 5 2
Now we take the resultant values and XOR them together:
3 ⊕ 4 ⊕ 5 ⊕ 7 ⊕ 1⊕ 2 = 6 we will return 6.
Problem Constraints
1 <= N <= 105
1 <= A[i] <= 108
Input Format
First and only argument is an integer array A.
Output Format
Return a single integer denoting the value as described above.
Example Input
Input 1:
A = [1, 2, 3]
Input 2:
A = [4, 5, 7, 5]
Example Output
Output 1:
2
Output 2:
0
Example Explanation
Explanation 1:
1 ⊕ 2 ⊕ 3 ⊕ (1 ⊕ 2 ) ⊕ (2 ⊕ 3) ⊕ (1 ⊕ 2 ⊕ 3) = 2
Explanation 2:
4 ⊕ 5 ⊕ 7 ⊕ 5 ⊕ (4 ⊕ 5) ⊕ (5 ⊕ 7) ⊕ (7 ⊕ 5) ⊕ (4 ⊕ 5 ⊕ 7) ⊕ (5 ⊕ 7 ⊕ 5) ⊕ (4 ⊕ 5 ⊕ 7 ⊕ 5) = 0
*/
public class Solution {
public int solve(ArrayList<Integer> A) {
//Like any element ith index in array A will occur (i + 1) * (N - i) times.
//for even size array all elements will exist even times
//for odd size array elements at (i+1)-> odd index will exist odd times others exist even times
int result = 0;
if(A.size() % 2 == 0){
return 0;
}
else{
for(int i=0; i<A.size(); i=i+2){
result = result ^ A.get(i);
}
}
return result;
}
}