def split_string_cost(S):
    # Length of the string S
    len_S = len(S)
   
    # To store the cost of the split parts
    max_cost = 0
   
    # Set to keep track of distinct characters in the first part
    distinct_chars_A = set()
   
    # List to keep track of the cost for the second part from each split position
    cost_B = [0] * len_S
   
    # Set to keep track of distinct characters in the second part
    distinct_chars_B = set()
   
    # Calculate cost for second part from the end
    for i in range(len_S - 1, -1, -1):
        distinct_chars_B.add(S[i])
        cost_B[i] = len(distinct_chars_B)
   
    # Calculate maximum sum of cost for parts A and B
    for i in range(len_S - 1):
        distinct_chars_A.add(S[i])
        cost_A = len(distinct_chars_A)
        cost = cost_A + cost_B[i + 1]
        max_cost = max(max_cost, cost)
   
    # Calculate the result as |S| - X
    result = len_S - max_cost
    return result

# Example usage
S = "aaabbb"
print(split_string_cost(S))  # Output: 3

#include <bits/stdc++.h>
using namespace std;

int equalzeroandone(vector<int>v){
    int n=v.size();
    for(int i=0;i<n;i++){
        if(v[i]==0){
            v[i]=-1;
        }
    }
    int sum=0;
    int ans=-1;
   map<int,int>mp;
    for(int i=0;i<n;i++){
        sum+=v[i];
        if(sum==0){
            ans=i+1;
        }
        if(mp.find(sum)!=mp.end()){
            ans=max(ans,i-mp[sum]);
        }
        else{
            mp[sum]=i;
        }
    }
    return ans;
}

int main() {
    int n;
    cin>>n;
    vector<int>v(n);
    for(int i=0;i<n;i++){
        cin>>v[i];
    }
    cout<<equalzeroandone(v);
}

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Attention everyone!!

Those who need help for Cognizant, Accenture, CGI or any placement tests clearance should contact @ILOVEU_143👨‍💻😊 and book your slots ✅

100% CLEARANCE ✅👨‍💻

Remote access available🤞

Telegram: @ILOVEU_143✅

SHARE @coding_000✅👨‍💻