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

Description

You are given two positive integers x and y.

In one operation, you can do one of the four following operations:

  1. Divide x by 11 if x is a multiple of 11.
  2. Divide x by 5 if x is a multiple of 5.
  3. Decrement x by 1.
  4. Increment x by 1.

Return the minimum number of operations required to make x and y equal.

 

Example 1:

Input: x = 26, y = 1
Output: 3
Explanation: We can make 26 equal to 1 by applying the following operations: 
1. Decrement x by 1
2. Divide x by 5
3. Divide x by 5
It can be shown that 3 is the minimum number of operations required to make 26 equal to 1.

Example 2:

Input: x = 54, y = 2
Output: 4
Explanation: We can make 54 equal to 2 by applying the following operations: 
1. Increment x by 1
2. Divide x by 11 
3. Divide x by 5
4. Increment x by 1
It can be shown that 4 is the minimum number of operations required to make 54 equal to 2.

Example 3:

Input: x = 25, y = 30
Output: 5
Explanation: We can make 25 equal to 30 by applying the following operations: 
1. Increment x by 1
2. Increment x by 1
3. Increment x by 1
4. Increment x by 1
5. Increment x by 1
It can be shown that 5 is the minimum number of operations required to make 25 equal to 30.

 

Constraints:

  • 1 <= x, y <= 104

Solutions

  • class Solution {
    private Map<Integer, Integer> f = new HashMap<>();
    private int y;

    public int minimumOperationsToMakeEqual(int x, int y) {
        this.y = y;
        return dfs(x);
    }

    private int dfs(int x) {
        if (y >= x) {
            return y - x;
        }
        if (f.containsKey(x)) {
            return f.get(x);
        }
        int ans = x - y;
        int a = x % 5 + 1 + dfs(x / 5);
        int b = 5 - x % 5 + 1 + dfs(x / 5 + 1);
        int c = x % 11 + 1 + dfs(x / 11);
        int d = 11 - x % 11 + 1 + dfs(x / 11 + 1);
        ans = Math.min(ans, Math.min(a, Math.min(b, Math.min(c, d))));
        f.put(x, ans);
        return ans;
    }
}

  • class Solution {
public:
    int minimumOperationsToMakeEqual(int x, int y) {
        unordered_map<int, int> f;
        function<int(int)> dfs = [&](int x) {
            if (y >= x) {
                return y - x;
            }
            if (f.count(x)) {
                return f[x];
            }
            int a = x % 5 + 1 + dfs(x / 5);
            int b = 5 - x % 5 + 1 + dfs(x / 5 + 1);
            int c = x % 11 + 1 + dfs(x / 11);
            int d = 11 - x % 11 + 1 + dfs(x / 11 + 1);
            return f[x] = min({x - y, a, b, c, d});
        };
        return dfs(x);
    }
};

  • class Solution:
    def minimumOperationsToMakeEqual(self, x: int, y: int) -> int:
        @cache
        def dfs(x: int) -> int:
            if y >= x:
                return y - x
            ans = x - y
            ans = min(ans, x % 5 + 1 + dfs(x // 5))
            ans = min(ans, 5 - x % 5 + 1 + dfs(x // 5 + 1))
            ans = min(ans, x % 11 + 1 + dfs(x // 11))
            ans = min(ans, 11 - x % 11 + 1 + dfs(x // 11 + 1))
            return ans

        return dfs(x)


  • func minimumOperationsToMakeEqual(x int, y int) int {
	f := map[int]int{}
	var dfs func(int) int
	dfs = func(x int) int {
		if y >= x {
			return y - x
		}
		if v, ok := f[x]; ok {
			return v
		}
		a := x%5 + 1 + dfs(x/5)
		b := 5 - x%5 + 1 + dfs(x/5+1)
		c := x%11 + 1 + dfs(x/11)
		d := 11 - x%11 + 1 + dfs(x/11+1)
		f[x] = min(x-y, a, b, c, d)
		return f[x]
	}
	return dfs(x)
}

  • function minimumOperationsToMakeEqual(x: number, y: number): number {
    const f: Map<number, number> = new Map();
    const dfs = (x: number): number => {
        if (y >= x) {
            return y - x;
        }
        if (f.has(x)) {
            return f.get(x)!;
        }
        const a = (x % 5) + 1 + dfs((x / 5) | 0);
        const b = 5 - (x % 5) + 1 + dfs(((x / 5) | 0) + 1);
        const c = (x % 11) + 1 + dfs((x / 11) | 0);
        const d = 11 - (x % 11) + 1 + dfs(((x / 11) | 0) + 1);
        const ans = Math.min(x - y, a, b, c, d);
        f.set(x, ans);
        return ans;
    };
    return dfs(x);
}

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