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 :
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
View More Similar Problems
Reverse a linked list
Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio
View Solution →Compare two linked lists
You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis
View Solution →Merge two sorted linked lists
This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C
View Solution →Get Node Value
This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t
View Solution →Delete duplicate-value nodes from a sorted linked list
This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -
View Solution →Cycle Detection
A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example head refers 1 -> 2 -> 3 -> NUL The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0. head refer
View Solution →