LeetCode #2844 — MEDIUM

Minimum Operations to Make a Special Number

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

Problem Statement

You are given a 0-indexed string num representing a non-negative integer.

In one operation, you can pick any digit of num and delete it. Note that if you delete all the digits of num, num becomes 0.

Return the minimum number of operations required to make num special.

An integer x is considered special if it is divisible by 25.

Example 1:

Input: num = "2245047"
Output: 2
Explanation: Delete digits num[5] and num[6]. The resulting number is "22450" which is special since it is divisible by 25.
It can be shown that 2 is the minimum number of operations required to get a special number.

Example 2:

Input: num = "2908305"
Output: 3
Explanation: Delete digits num[3], num[4], and num[6]. The resulting number is "2900" which is special since it is divisible by 25.
It can be shown that 3 is the minimum number of operations required to get a special number.

Example 3:

Input: num = "10"
Output: 1
Explanation: Delete digit num[0]. The resulting number is "0" which is special since it is divisible by 25.
It can be shown that 1 is the minimum number of operations required to get a special number.

Constraints:

1 <= num.length <= 100
num only consists of digits '0' through '9'.
num does not contain any leading zeros.

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed string num representing a non-negative integer. In one operation, you can pick any digit of num and delete it. Note that if you delete all the digits of num, num becomes 0. Return the minimum number of operations required to make num special. An integer x is considered special if it is divisible by 25.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Greedy

Example 1

"2245047"

Example 2

"2908305"

Example 3

"10"

Core Insight

What unlocks the optimal approach

If <code>num</code> contains a single zero digit then the answer is at most <code>n - 1</code>.
A number is divisible by <code>25</code> if its last two digits are <code>75</code>, <code>50</code>, <code>25</code>, or <code>00</code>.
Iterate over all possible pairs of indices <code>i < j</code> such that <code>num[i] * 10 + num[j]</code> is in <code>[00,25,50,75]</code>. Then, set the answer to <code> min(answer, n - i - 2) </code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2844: Minimum Operations to Make a Special Number
class Solution {
    private Integer[][] f;
    private String num;
    private int n;

    public int minimumOperations(String num) {
        n = num.length();
        this.num = num;
        f = new Integer[n][25];
        return dfs(0, 0);
    }

    private int dfs(int i, int k) {
        if (i == n) {
            return k == 0 ? 0 : n;
        }
        if (f[i][k] != null) {
            return f[i][k];
        }
        f[i][k] = dfs(i + 1, k) + 1;
        f[i][k] = Math.min(f[i][k], dfs(i + 1, (k * 10 + num.charAt(i) - '0') % 25));
        return f[i][k];
    }
}

// Accepted solution for LeetCode #2844: Minimum Operations to Make a Special Number
func minimumOperations(num string) int {
	n := len(num)
	f := make([][25]int, n)
	for i := range f {
		for j := range f[i] {
			f[i][j] = -1
		}
	}
	var dfs func(i, k int) int
	dfs = func(i, k int) int {
		if i == n {
			if k == 0 {
				return 0
			}
			return n
		}
		if f[i][k] != -1 {
			return f[i][k]
		}
		f[i][k] = dfs(i+1, k) + 1
		f[i][k] = min(f[i][k], dfs(i+1, (k*10+int(num[i]-'0'))%25))
		return f[i][k]
	}
	return dfs(0, 0)
}

# Accepted solution for LeetCode #2844: Minimum Operations to Make a Special Number
class Solution:
    def minimumOperations(self, num: str) -> int:
        @cache
        def dfs(i: int, k: int) -> int:
            if i == n:
                return 0 if k == 0 else n
            ans = dfs(i + 1, k) + 1
            ans = min(ans, dfs(i + 1, (k * 10 + int(num[i])) % 25))
            return ans

        n = len(num)
        return dfs(0, 0)

// Accepted solution for LeetCode #2844: Minimum Operations to Make a Special Number
// 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 #2844: Minimum Operations to Make a Special Number
// class Solution {
//     private Integer[][] f;
//     private String num;
//     private int n;
// 
//     public int minimumOperations(String num) {
//         n = num.length();
//         this.num = num;
//         f = new Integer[n][25];
//         return dfs(0, 0);
//     }
// 
//     private int dfs(int i, int k) {
//         if (i == n) {
//             return k == 0 ? 0 : n;
//         }
//         if (f[i][k] != null) {
//             return f[i][k];
//         }
//         f[i][k] = dfs(i + 1, k) + 1;
//         f[i][k] = Math.min(f[i][k], dfs(i + 1, (k * 10 + num.charAt(i) - '0') % 25));
//         return f[i][k];
//     }
// }

// Accepted solution for LeetCode #2844: Minimum Operations to Make a Special Number
function minimumOperations(num: string): number {
    const n = num.length;
    const f: number[][] = Array.from({ length: n }, () => Array.from({ length: 25 }, () => -1));
    const dfs = (i: number, k: number): number => {
        if (i === n) {
            return k === 0 ? 0 : n;
        }
        if (f[i][k] !== -1) {
            return f[i][k];
        }
        f[i][k] = dfs(i + 1, k) + 1;
        f[i][k] = Math.min(f[i][k], dfs(i + 1, (k * 10 + Number(num[i])) % 25));
        return f[i][k];
    };
    return dfs(0, 0);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × 25)

Space

O(n × 25)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.