leetcode-solution/991.broken-calculator.cpp at master · oddway/leetcode-solution · GitHub

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/*
 * @lc app=leetcode id=991 lang=cpp
 *
 * [991] Broken Calculator
 *
 * https://leetcode.com/problems/broken-calculator/description/
 *
 * algorithms
 * Medium (45.04%)
 * Likes:    410
 * Dislikes: 83
 * Total Accepted:    16.8K
 * Total Submissions: 37.2K
 * Testcase Example:  '2\n3'
 *
 * On a broken calculator that has a number showing on its display, we can
 * perform two operations:
 *
 *
 * Double: Multiply the number on the display by 2, or;
 * Decrement: Subtract 1 from the number on the display.
 *
 *
 * Initially, the calculator is displaying the number X.
 *
 * Return the minimum number of operations needed to display the number Y.
 *
 *
 *
 * Example 1:
 *
 *
 * Input: X = 2, Y = 3
 * Output: 2
 * Explanation: Use double operation and then decrement operation {2 -> 4 ->
 * 3}.
 *
 *
 * Example 2:
 *
 *
 * Input: X = 5, Y = 8
 * Output: 2
 * Explanation: Use decrement and then double {5 -> 4 -> 8}.
 *
 *
 * Example 3:
 *
 *
 * Input: X = 3, Y = 10
 * Output: 3
 * Explanation:  Use double, decrement and double {3 -> 6 -> 5 -> 10}.
 *
 *
 * Example 4:
 *
 *
 * Input: X = 1024, Y = 1
 * Output: 1023
 * Explanation: Use decrement operations 1023 times.
 *
 *
 *
 *
 * Note:
 *
 *
 * 1 <= X <= 10^9
 * 1 <= Y <= 10^9
 *
 */

// @lc code=start
class Solution {
   public:
    int brokenCalc(int X, int Y) {
        int ans = 0;

        while (Y > X) {
            ans++;
            if (Y % 2 == 1) {
                Y++;
            } else {
                Y /= 2;
            }
        }

        return ans + X - Y;
    }

    // recursive method
    /* int brokenCalc(int X, int Y) {
        if (X >= Y)
        {
            return X - Y;
        }

        int opCounter = 0;

        if (Y % 2 != 0)
        {
            Y++;
            opCounter++;
        }

        Y /= 2;
        opCounter++;

        opCounter += brokenCalc(X, Y);

        return opCounter;
    } */
};
// @lc code=end