Maximal Points From Deleting Two Character Substrings- Google Top Interview Questions


Problem Statement :


You are given a string s containing "1"s and "0"s and integers zeroone and onezero. 

In one operation you can remove any substring "01" and receive zeroone points.

 Or you can remove any substring "10" and receive onezero points.

Return the maximum number of points you can receive, given that you can make any number of operations.

Constraints

n ≤ 100,000 where n is the length of s

Example 1

Input

s = "101010"

zeroone = 3

onezero = 1

Output

7

Explanation

We can delete "01" twice to receive 3 points each. The resulting string then becomes "10" and then you 
can delete it to receive 1 point.


Solution :



title-img



                        Solution in C++ :

int solve(string s, int zeroone, int onezero) {
    bool flag = 0;
    if (zeroone > onezero) flag = 1;
    stack<char> st;
    int ans = 0;
    for (int i = 0; i < s.size(); i++) {
        if (st.empty()) {
            st.push(s[i]);
            continue;
        }
        if (s[i] == '0' && st.top() == '1' && !flag) {
            ans += onezero;
            st.pop();
        } else if (s[i] == '1' && st.top() == '0' && flag) {
            ans += zeroone;
            st.pop();
        } else
            st.push(s[i]);
    }
    int one = 0, zero = 0;
    while (!st.empty()) {
        if (st.top() == '0') zero++;
        if (st.top() == '1') one++;
        st.pop();
    }
    if (flag)
        ans += (onezero * min(zero, one));
    else
        ans += (zeroone * min(zero, one));
    return ans;
}
                    

                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(String s, int A, int B) {
        int N = s.length();
        int[] nums = new int[N];
        int X = 0;
        for (int i = 0; i < N; i++) {
            nums[i] = s.charAt(i) - '0';
            if (nums[i] == 0)
                X++;
        }
        X = Math.min(X, N - X);
        if (A < B) {
            int temp = A;
            A = B;
            B = temp;
            for (int i = 0; i < N; i++) {
                nums[i] = 1 - nums[i];
            }
        }
        int zero = 0;
        int get = 0;
        for (int i = 0; i < N; i++) {
            if (nums[i] == 0) {
                zero++;
            } else {
                if (zero > 0) {
                    zero--;
                    get++;
                }
            }
        }
        int ans = get * A + (X - get) * B;
        return ans;
    }
}
                    

                        Solution in Python : 
                            
class Solution:
    def solve(self, S, pts01, pts10):
        A = list(map(int, S))
        if pts01 < pts10:
            pts01, pts10 = pts10, pts01
            for i in range(len(A)):
                A[i] ^= 1

        ans = 0
        stack = []
        for x in A:
            if stack and stack[-1] < x:
                stack.pop()
                ans += pts01
            else:
                stack.append(x)

        ans += pts10 * min(stack.count(0), stack.count(1))
        return ans
                    

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