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XORQueriesOfASubarray1310.java
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XORQueriesOfASubarray1310.java
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import java.util.* ;
import java.util.ArrayList ;
public class XORQueriesOfASubarray1310 {
/*
* You are given an array arr of positive integers. You are also given the array queries where queries[i] = [lefti, righti].
For each query i compute the XOR of elements from lefti to righti (that is, arr[lefti] XOR arr[lefti + 1] XOR ... XOR arr[righti] ).
Return an array answer where answer[i] is the answer to the ith query.
*/
class Solution {
public int[] xorQueries(int[] arr, int[][] queries) {
int ans[] = new int[queries.length] ;
for (int i = 0; i < queries.length; i++) {
int req = 0 ;
for (int j = queries[i][0] ; j<= queries[i][1] ; j++){
req ^= arr[j];
}
ans[i] = req;
}
return ans;
}
}
//minimize the time by increasing space
class Solution1 {
public int[] xorQueries(int[] arr, int[][] queries) {
int n= arr.length;
int [] allQueryAns = new int [n+ 1] ;
for (int i = 0; i < n; i++)
allQueryAns[i+1] = allQueryAns[i] ^ arr[i];
int [] ans = new int[queries.length] ;
for (int i = 0; i < queries.length; i++) {
ans[i] = allQueryAns[queries[i][1]+1] ^ allQueryAns[queries[i][0]];
}
return ans;
}
}
////minimize the time by without increasing space
class Solution3 {
public int[] xorQueries(int[] arr, int[][] queries) {
List<Integer> result = new ArrayList<>();
for (int i = 1; i < arr.length; ++i)
arr[i] ^= arr[i - 1];
for (int[] q : queries) {
if (q[0] > 0) result.add(arr[q[0] - 1] ^ arr[q[1]]);
else result.add(arr[q[1]]);
}
return result.stream().mapToInt(i -> i).toArray();
}
}
}