allcoding1
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Removable subarrays
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Copy ur question and past you get Answer
#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;

const int MOD = 1e9 + 7;

vector<vector<int>> tree;
vector<int> a;
unordered_map<int, int> countMap;

bool checkPalindrome(unordered_map<int, int>& countMap) {
int oddCount = 0;
for (auto& it : countMap) {
if (it.second % 2 != 0) oddCount++;
if (oddCount > 1) return false;
}
return true;
}

int dfs(int node) {
int ans = 0;
countMap[a[node]]++;

if (checkPalindrome(countMap)) {
ans = 1;
}

for (auto& child : tree[node]) {
ans += dfs(child);
ans %= MOD;
}

countMap[a[node]]--; // Backtrack to remove the current node's count
return ans;
}

int main() {
int n;
cin >> n;

tree.resize(n + 1);
a.resize(n + 1);
vector<int> par(n + 1);

for (int i = 2; i <= n; i++) {
cin >> par[i];
tree[par[i]].push_back(i);
}

for (int i = 1; i <= n; i++) {
cin >> a[i];
}

int ans = dfs(1);
cout << ans << endl;

return 0;
}



c++

Palindromic subtrees

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#include <iostream>
#include <string>

using namespace std;

bool isGoodString(string s) {
    int count = 0;
    for (char c : s) {
        if (c == 'a') {
            count++;
        } else {
            count--;
        }
        if (count < 0) {
            return false;
        }
    }
    return count == 0;
}

int main() {
    string s = "aaabbbaaa";
    cout << "Is the string good? " << (isGoodString(s) ? "Yes" : "No") << endl;
    return 0;
}

Count Good Triples
C++
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