2844. Minimum Operations to Make a Special Number

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.

Intuition

Approach 1:
DP

Take/ Not Take

Not able to solve in the contest, Because passing string as parameter, Rather 
Than passing number

And taking minimum

Approach 2:
Number divisible by 25, when last two nums are 00,25,50,75
Then take two pointers and iterate

Take care of one 0 condition, in that case we delete all

Links

https://leetcode.com/problems/minimum-operations-to-make-a-special-number/description/

Video Links

https://www.youtube.com/watch?v=sHFB71qC4wE&ab_channel=AryanMittal

Approach 1:

DP

C++

class Solution {
public:
    vector<vector<int>> dp;
    int find_ans(string &s, int index, int temp){
        if(index == s.size()){
            if(temp % 25 == 0)  return 0;
            return 1e9+7;
        }

        if(dp[index][temp] != -1)
            return dp[index][temp];
        
        int not_del = 1 + find_ans(s, index+1, temp);

        int digit=s[index]-'0';
        int del = find_ans(s, index+1, (temp*10+digit)%25);

        return dp[index][temp] = min(del, not_del);
    }

    int minimumOperations(string num) {
        dp = vector<vector<int>> (102, vector<int>(27,-1));
        return find_ans(num, 0, 0);
    }
};

Approach 2:

Looping

C++

class Solution {
public:
    int minimumOperations(string num) {
        int n=num.size();
        int ans=num.size();
        // Worst case all deletion

        for(int i=n-1; i>=0; i--){
            for(int j=i-1; j>=0; j--){
                int temp = (num[j]-'0')*10 + (num[i]-'0');
                if(temp %25 == 0)
                    ans = min(ans, n-j-2);
            }
            if(num[i] == '0')
                ans = min(ans, n-1);
        }

        return ans;
    }
};

Approach 3:

C++

Approach 4:

C++

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