XOR Queries of a Subarray

XOR Queries of a Subarray - Problem

You are given an array arr of positive integers. You are also given the array queries where queries[i] = [left_i, right_i].

For each query i, compute the XOR of elements from left_i to right_i (that is, arr[left_i] XOR arr[left_i + 1] XOR ... XOR arr[right_i]).

Return an array answer where answer[i] is the answer to the ith query.

Input & Output

Example 1 — Basic Range Queries

$ Input: arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]]

› Output: [2,7,14,8]

💡 Note: Query [0,1]: 1 ^ 3 = 2. Query [1,2]: 3 ^ 4 = 7. Query [0,3]: 1 ^ 3 ^ 4 ^ 8 = 14. Query [3,3]: 8 = 8.

Example 2 — Single Element Query

$ Input: arr = [4,8,2,10], queries = [[2,3],[1,3],[0,0],[0,3]]

› Output: [8,4,4,8]

💡 Note: Query [2,3]: 2 ^ 10 = 8. Query [1,3]: 8 ^ 2 ^ 10 = 4. Query [0,0]: 4 = 4. Query [0,3]: 4 ^ 8 ^ 2 ^ 10 = 8.

Example 3 — Edge Case

$ Input: arr = [5], queries = [[0,0]]

› Output: [5]

💡 Note: Single element array with single element query returns the element itself.

Constraints

1 ≤ arr.length ≤ 3 × 10⁴
1 ≤ arr[i] ≤ 10⁹
1 ≤ queries.length ≤ 3 × 10⁴
queries[i].length == 2
0 ≤ left_i ≤ right_i < arr.length

Visualization

Tap to expand

Asked in

a Amazon 25 M Microsoft 18

The key insight is that XOR has a special property: A ^ B ^ B = A, which allows us to use prefix XOR arrays. The optimal Prefix XOR approach precomputes cumulative XOR values, then answers any range query in O(1) time using the formula: XOR(left, right) = prefix[right+1] ^ prefix[left]. Time: O(n + m), Space: O(n).

Common Approaches

✓ Brute Force

⏱️ Time: O(m × n) Space: O(1)

For every query, we iterate from left to right index and XOR all elements in that range. This is straightforward but inefficient for multiple queries.

Prefix XOR

⏱️ Time: O(n + m) Space: O(n)

We precompute prefix XOR array where prefix[i] contains XOR of all elements from 0 to i. Then any range XOR can be computed using the property: XOR(left, right) = prefix[right] ^ prefix[left-1].

Brute Force — Algorithm Steps

For each query [left, right]
Iterate from arr[left] to arr[right]
XOR all elements in the range
Store result in answer array

Visualization

Tap to expand

Step-by-Step Walkthrough

Query [1,3]

XOR elements at indices 1 to 3

Compute

arr[1] ^ arr[2] ^ arr[3] = 4 ^ 3 ^ 5 = 6

Result

Store 6 in answer array

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    
    // Find first '['
    while (*p && *p != '[') p++;
    if (!*p) return;
    p++; // skip '['
    
    char num[20];
    int numIdx = 0;
    
    while (*p && *p != ']') {
        if (isdigit(*p) || *p == '-') {
            num[numIdx++] = *p;
        } else if (*p == ',' && numIdx > 0) {
            num[numIdx] = '\0';
            arr[(*size)++] = atoi(num);
            numIdx = 0;
        }
        p++;
    }
    
    if (numIdx > 0) {
        num[numIdx] = '\0';
        arr[(*size)++] = atoi(num);
    }
}

void parseQueries(const char* str, int queries[][2], int* queriesSize) {
    *queriesSize = 0;
    const char* p = str;
    
    // Find first '['
    while (*p && *p != '[') p++;
    if (!*p) return;
    p++; // skip first '['
    
    while (*p) {
        // Find start of next sub-array
        while (*p && *p != '[') p++;
        if (!*p) break;
        p++; // skip '['
        
        char num[20];
        int numIdx = 0;
        int queryIdx = 0;
        
        // Parse numbers until we hit ']'
        while (*p && *p != ']' && queryIdx < 2) {
            if (isdigit(*p) || *p == '-') {
                num[numIdx++] = *p;
            } else if (*p == ',' && numIdx > 0) {
                num[numIdx] = '\0';
                queries[*queriesSize][queryIdx++] = atoi(num);
                numIdx = 0;
            }
            p++;
        }
        
        if (numIdx > 0 && queryIdx < 2) {
            num[numIdx] = '\0';
            queries[*queriesSize][queryIdx++] = atoi(num);
        }
        
        if (queryIdx == 2) {
            (*queriesSize)++;
        }
        
        if (*p == ']') p++; // skip ']'
    }
}

int* solution(int* arr, int arrSize, int queries[][2], int queriesSize, int* returnSize) {
    int* answer = (int*)malloc(queriesSize * sizeof(int));
    *returnSize = queriesSize;
    
    for (int i = 0; i < queriesSize; i++) {
        int left = queries[i][0];
        int right = queries[i][1];
        int xor_result = 0;
        for (int j = left; j <= right; j++) {
            xor_result ^= arr[j];
        }
        answer[i] = xor_result;
    }
    
    return answer;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    static int arr[1000];
    int arrSize;
    parseArray(line, arr, &arrSize);
    
    fgets(line, sizeof(line), stdin);
    
    // Parse queries
    static int queries[100][2];
    int queriesSize;
    parseQueries(line, queries, &queriesSize);
    
    int returnSize;
    int* result = solution(arr, arrSize, queries, queriesSize, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("%d", result[i]);
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(m × n)

m queries, each taking O(n) time to iterate through array

✓ Linear Growth

Space Complexity

O(1)

Only using variables for current XOR calculation

⚡ Linearithmic Space

28.4K Views

Medium Frequency

~15 min Avg. Time

892 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler