Keep Multiplying Found Values by Two

Keep Multiplying Found Values by Two - Problem

You are given an array of integers nums. You are also given an integer original which is the first number that needs to be searched for in nums.

You then do the following steps:

If original is found in nums, multiply it by two (i.e., set original = 2 * original).
Otherwise, stop the process.
Repeat this process with the new number as long as you keep finding the number.

Return the final value of original.

Input & Output

Example 1 — Basic Case

$ Input: nums = [5,3,6,1,12], original = 3

› Output: 24

💡 Note: 3 is found → double to 6. 6 is found → double to 12. 12 is found → double to 24. 24 is not found → return 24

Example 2 — Single Step

$ Input: nums = [2,7,9], original = 4

› Output: 4

💡 Note: 4 is not found in the array, so return the original value 4

Example 3 — Multiple Doublings

$ Input: nums = [1,2,4,8], original = 1

› Output: 16

💡 Note: 1 → 2 → 4 → 8 → 16. Since 16 is not in array, return 16

Constraints

1 ≤ nums.length ≤ 1000
1 ≤ nums[i], original ≤ 1000

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to use a hash set for O(1) lookups instead of repeatedly searching the array. Convert the array to a set, then keep doubling the value while it exists in the set. Best approach is hash set: Time: O(n + k), Space: O(n)

Common Approaches

✓ Brute Force Linear Search

⏱️ Time: O(n × k) Space: O(1)

For each iteration, scan through the entire array to check if the current value exists. If found, double it and repeat. Continue until the doubled value is not found in the array.

Hash Set Optimization

⏱️ Time: O(n + k) Space: O(n)

Convert the array to a hash set once, then perform O(1) lookups for each doubled value. This eliminates the need to scan the entire array repeatedly.

Brute Force Linear Search — Algorithm Steps

Start with original value
Scan entire array to check if current value exists
If found, double the value and repeat
If not found, return current value

Visualization

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Step-by-Step Walkthrough

Search Array

Scan entire array looking for current value

Double Value

If found, multiply by 2 and repeat

Stop

When value not found, return result

Code -

solution.c — C

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

int solution(int* nums, int numsSize, int original) {
    int current = original;
    while (1) {
        int found = 0;
        for (int i = 0; i < numsSize; i++) {
            if (nums[i] == current) {
                found = 1;
                break;
            }
        }
        if (found) {
            current *= 2;
        } else {
            break;
        }
    }
    return current;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int original;
    scanf("%d", &original);
    
    int result = solution(nums, numsSize, original);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × k)

For each of k doublings, we scan the array of n elements

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables, no extra data structures

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force Linear Search — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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