Construct the Minimum Bitwise Array I

Construct the Minimum Bitwise Array I - Problem

You are given an array nums consisting of n prime integers. You need to construct an array ans of length n, such that for each index i, the bitwise OR of ans[i] and ans[i] + 1 is equal to nums[i].

In other words: ans[i] OR (ans[i] + 1) == nums[i]

Additionally, you must minimize each value of ans[i] in the resulting array. If it is not possible to find such a value for ans[i] that satisfies the condition, then set ans[i] = -1.

Input & Output

Example 1 — Basic Case

$ Input: nums = [2,3,5,7]

› Output: [-1,1,4,3]

💡 Note: For 2: no valid x exists since 2=10 in binary. For 3: 1|2=3. For 5: 4|5=5. For 7: 3|4=7.

Example 2 — Larger Numbers

$ Input: nums = [11,13,31]

› Output: [10,12,30]

💡 Note: For 11: 10|11=11. For 13: 12|13=13. For 31: 30|31=31.

Example 3 — Mixed Valid and Invalid

$ Input: nums = [2,5,7]

› Output: [-1,4,3]

💡 Note: 2 has no solution, 5 maps to 4, and 7 maps to 3.

Constraints

1 ≤ nums.length ≤ 100
2 ≤ nums[i] ≤ 10⁹
nums[i] is a prime number

Visualization

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Asked in

G Google 15 M Microsoft 12 a Amazon 8

The key insight is understanding how bitwise OR works with consecutive numbers: x | (x+1) sets the rightmost 0 bit in x and all bits to its right. The optimal approach uses bit manipulation to directly calculate candidates using num & (num-1), then verify the result. Time: O(n), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force Search	O(n × max_value)	O(1)	Try all possible values for each ans[i] until finding one that satisfies the OR condition
Bit Manipulation Analysis	O(n)	O(1)	Analyze bit patterns to directly find the answer or determine impossibility

Brute Force Search — Algorithm Steps

For each nums[i], start with x = 0
Check if x OR (x+1) equals nums[i]
If yes, add x to result; if no, increment x and repeat
If no valid x found within bounds, add -1

Visualization

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Step-by-Step Walkthrough

Start Search

For nums[i], try x = 0, 1, 2, ...

Check Condition

Test if x | (x+1) == nums[i]

Found or Continue

First match is minimum, or -1 if none

Code -

solution.c — C

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

int* solution(int* nums, int numsSize, int* returnSize) {
    int* result = (int*)malloc(numsSize * sizeof(int));
    *returnSize = numsSize;
    
    for (int i = 0; i < numsSize; i++) {
        int num = nums[i];
        int found = 0;
        
        // Try values up to num
        for (int x = 0; x <= num; x++) {
            if ((x | (x + 1)) == num) {
                result[i] = x;
                found = 1;
                break;
            }
        }
        
        if (!found) {
            result[i] = -1;
        }
    }
    
    return result;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Simple JSON array parsing
    int nums[1000];
    int count = 0;
    char* token = strtok(line, "[],\n");
    
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != ' ') {
            nums[count++] = atoi(token);
        }
        token = strtok(NULL, "[],\n");
    }
    
    int returnSize;
    int* result = solution(nums, count, &returnSize);
    
    // Print result
    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(n × max_value)

For each element, we might check up to max_value possibilities

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space for variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force Search — Algorithm Steps

Visualization

Code -

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