LeetCode 2998: Minimum Number of Operations to Make X and Y Equal Solution

Master LeetCode problem 2998 (Minimum Number of Operations to Make X and Y Equal), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2998. Minimum Number of Operations to Make X and Y Equal

Medium

Dynamic Programming Breadth-First Search Memoization

Problem Explanation

Explanation

To solve this problem, we can work backwards from the target value y to x. Since the operations are limited to dividing by 11, dividing by 5, incrementing, and decrementing, we can consider the possible last steps to reach y from x. We can iterate through all possible last steps and recursively find the minimum number of operations required to reach each previous state.

The algorithm involves keeping track of the minimum operations required to reach each state. We can use a dynamic programming approach to store the minimum operations for each state.

Initialize a map to store the minimum operations for each state.
Start from y and work backwards to x.
For each state, consider all possible last steps (decrement, increment, divide by 5, divide by 11) and recursively find the minimum operations required.
Store the minimum operations for each state in the map.
Finally, return the minimum operations required to reach x from y.

Time Complexity: The time complexity of this algorithm is O(xy) since we are considering all possible states between x and y.

Space Complexity: The space complexity is also O(xy) due to the dynamic programming map storing the minimum operations for each state.

Solution Code

import java.util.*;

class Solution {
    public int minOperations(int x, int y) {
        Map<Integer, Integer> dp = new HashMap<>();
        int result = helper(x, y, dp);
        return result == Integer.MAX_VALUE ? -1 : result;
    }

    private int helper(int x, int y, Map<Integer, Integer> dp) {
        if (x == y) {
            return 0;
        }
        if (x == 0 || x > y) {
            return Integer.MAX_VALUE;
        }
        if (dp.containsKey(x)) {
            return dp.get(x);
        }
        int minOps = Integer.MAX_VALUE;
        if (y % 2 == 0) {
            minOps = Math.min(minOps, 1 + helper(x, y / 2, dp));
        }
        if (y % 3 == 0) {
            minOps = Math.min(minOps, 1 + helper(x, y / 3, dp));
        }
        minOps = Math.min(minOps, 1 + helper(x, y - 1, dp));
        minOps = Math.min(minOps, 1 + helper(x, y + 1, dp));
        dp.put(x, minOps);
        return minOps;
    }
}

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