leetcode2154: Keep Multiplying Found Values by Two

[carloscn]

Description

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.

Example 1:

Input: nums = [5,3,6,1,12], original = 3 Output: 24 Explanation:

• 3 is found in nums. 3 is multiplied by 2 to obtain 6.
• 6 is found in nums. 6 is multiplied by 2 to obtain 12.
• 12 is found in nums. 12 is multiplied by 2 to obtain 24.
• 24 is not found in nums. Thus, 24 is returned.

Example 2:

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

• 4 is not found in nums. Thus, 4 is returned.

Constraints:

1 <= nums.length <= 1000 1 <= nums[i], original <= 1000

[carloscn]

Analysis

int32_t find_final_value(int32_t* nums, size_t nums_size, int32_t original, int32_t *out)
{
    int32_t ret = 0;

    UTILS_CHECK_PTR(nums);
    UTILS_CHECK_PTR(out);
    UTILS_CHECK_LEN(nums_size);

    for (size_t i = 0; i < nums_size; i ++) {
        for (size_t j = 0; j < nums_size - i - 1; j ++) {
            if (nums[j] > nums[j + 1]) {
                utils_swap_int32(nums + j, nums + j + 1);
            }
        }
    }

    bool flag = false;
    size_t i = 0;
    while (i < nums_size) {
        if (nums[i] == original) {
            original *= 2;
            flag = true;
        }
        i ++;
    }

    *out = flag == true ? original * 2 : original;

    LOG("the output is %d\n", *out);

finish:
    return ret;
}

[carloscn]

Code

https://review.gerrithub.io/c/carloscn/structstudy/+/1169095 https://github.com/carloscn/structstudy/commit/980c682648b68735457dcff6f5b20a9b67c989a4