There are many ways to solve this problem. One of which is described here.

All we have to do is find the maximum number of indices such that both strings have different values. We can do this by sorting one string in increasing order and another one in decreasing order. Now count of these indices gives us the number of set bits and the remaining are unset bits.

Time Complexity: $$$O(n \log n)$$$

#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
int main() {
	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	
	string s, t;
	cin >> s >> t;
	sort(s.begin(), s.end());
	sort(t.begin(), t.end());
	reverse(t.begin(), t.end());
	string ans = "";
	for(int i = 0; i < (int) s.size(); i ++) {
		if(s[i] != t[i]) ans += "1";
		else ans += "0";
	}
	sort(ans.begin(), ans.end());
	reverse(ans.begin(), ans.end());
	cout << ans;
	
	return 0;
}

Go for the tutorial of GS more like BS(Hard version)

Like in Kadane's Algorithm, we set our max_ending_here variable to zero if and only if we are on good index and max_ending_here is less than zero.

Time Complexity: $$$O(n)$$$

#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
int main() {
	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	int n;  cin >> n;
        vector<int> arr(n), g(n);
        for(int i = 0; i < n; i ++)  cin >> arr[i];
        for(int i = 0; i < n; i ++)  cin >> g[i];
        ll max_sum_so_far = 0, max_sum = -2e18, st = -1;
        for(int i = 0; i < n; i ++) {
      	    if(g[i] == 1)	st = 1;
            if(g[i] == 1 and max_sum_so_far < 0)
                max_sum_so_far = 0;
            if(st == -1)	continue;
            max_sum_so_far += arr[i];
            max_sum = max(max_sum, max_sum_so_far);
        }
        cout << max_sum << endl;
	return 0;
}

Time Complexity: $$$O()$$$

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Time Complexity: $$$O()$$$

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Time Complexity: $$$O(n)$$$

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Time Complexity: $$$O()$$$

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