Move from brute-force thinking to an efficient approach using dynamic programming strategy.
You are given two positive integers x and y.
In one operation, you can do one of the four following operations:
x by 11 if x is a multiple of 11.x by 5 if x is a multiple of 5.x by 1.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 <= 104Problem summary: You are given two positive integers x and y. In one operation, you can do one of the four following operations: Divide x by 11 if x is a multiple of 11. Divide x by 5 if x is a multiple of 5. Decrement x by 1. Increment x by 1. Return the minimum number of operations required to make x and y equal.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Dynamic Programming
26 1
54 2
25 30
shortest-bridge)minimum-moves-to-spread-stones-over-grid)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2998: Minimum Number of Operations to Make X and Y Equal
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;
}
}
// Accepted solution for LeetCode #2998: Minimum Number of Operations to Make X and Y Equal
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)
}
# Accepted solution for LeetCode #2998: Minimum Number of Operations to Make X and Y Equal
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)
// Accepted solution for LeetCode #2998: Minimum Number of Operations to Make X and Y Equal
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
pub fn rust_example() {
// Port the logic from the reference block below.
}
}
// Reference (java):
// // Accepted solution for LeetCode #2998: Minimum Number of Operations to Make X and Y Equal
// 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;
// }
// }
// Accepted solution for LeetCode #2998: Minimum Number of Operations to Make X and Y Equal
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);
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.