Suppose we have a string with lowercase letters and we also have a list of non-negative values called costs, the string and the list have the same length. We can delete character s[i] for cost costs[i], and then both s[i] and costs[i] is removed. We have to find the minimum cost to delete all consecutively repeating characters.
So, if the input is like s = "xxyyx" nums = [2, 3, 10, 4, 6], then the output will be 6, as we can delete s[0] and s[3] for a total cost of 2 + 4 = 6.
To solve this, we will follow these steps
Define one stack st
cost := 0
for initialize i := 0, when i < size of s, update (increase i by 1), do:
if size of st is not 0 and s[top of st] is same as s[i], then:
if nums[top of st] > nums[i], then:
cost := cost + nums[i]
otherwise:
cost := cost + nums[top of st]
pop element from st
push i into st
otherwise
push i into st
return cost
Let us see the following implementation to get better understanding
Example
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int solve(string s, vector<int>& nums) {
stack<int> st;
int cost = 0;
for (int i = 0; i < s.size(); ++i) {
if (st.size() && s[st.top()] == s[i]) {
if (nums[st.top()] > nums[i]) {
cost += nums[i];
} else {
cost += nums[st.top()];
st.pop();
st.push(i);
}
} else {
st.push(i);
}
}
return cost;
}
};
int solve(string s, vector<int>& nums) {
return (new Solution())->solve(s, nums);
}
main(){
vector<int> v = {2, 3, 10, 4, 6};
string s = "xxyyx";
cout << solve(s, v);
}Input
"xxyyx",{2,3,10,4,6}Output
6