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

int minimumCost(int N, vector<int> A, vector<vector<int>> Cost) {
    int minCost = INT_MAX;
    vector<int> perm(N);
    iota(perm.begin(), perm.end(), 1);

    do {
        bool valid = true;
        for (int i = 0; i < N; ++i) {
            if (perm[i] > A[i]) {
                valid = false;
                break;
            } else if (perm[i] < A[i]) {
                break;
            }
        }

        if (valid) {
            int cost = 0;
            for (int i = 0; i < N; ++i) {
                cost += Cost[i][perm[i] - 1];
            }
            minCost = min(minCost, cost);
        }
    } while (next_permutation(perm.begin(), perm.end()));

    return minCost;
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int N;
    cin >> N;
    vector<int> A(N);
    for (int i = 0; i < N; ++i) {
        cin >> A[i];
    }

    vector<vector<int>> Cost(N, vector<int>(N));
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) {
            cin >> Cost[i][j];
        }
    }

    cout << minimumCost(N, A, Cost) << endl;

    return 0;
}

Cheapest permutation ✅

import java.util.*;

class Main {
  public static void main (String[] args) {
    Scanner sc=new Scanner(System.in);
    int n=sc.nextInt(),i;
    TreeMap<Long,Integer> map=new TreeMap<>();
    long dp[]=new long[n];
    long r[]=new long[n];
    long h[]=new long[n];
    long mod=(long)1e9+7;
    for(i=0;i<n;i++) r[i]=sc.nextLong();
    for(i=0;i<n;i++) h[i]=sc.nextLong();
    for(i=0;i<n;i++){
        long volume=(h[i]*r[i]*r[i])%mod;
        Long low=map.floorKey(volume);
        if(low!=null) dp[i]=(dp[i]+dp[map.get(low)]+volume)%mod;
        else dp[i]=(dp[i]+volume)%mod;
        map.put(volume,i);
    }
    long max=0;
    for(long it:dp) max=Math.max(max,it);
    System.out.println(max);
  }
}

Increasing mex counting ✅

#include <bits/stdc++.h>
using namespace std;
#define MOD 1000000007

vector<int> A, P;
vector<vector<int>> tree;
vector<int> ranges;
long long result = 0;

void dfs(int node, int parent, set<int>& values) {
    values.insert(A[node]);
    set<int> child_values;
    int num_ranges = 1, last = -1;

    for (int val : values) {
        if (last != -1 && val != last + 1) num_ranges++;
        last = val;
    }

    ranges[node] = num_ranges;

    for (int child : tree[node]) {
        if (child == parent) continue;
        child_values = values;
        dfs(child, node, child_values);
    }

    result = (result + ranges[node]) % MOD;
}

int main() {
    int N;
    cin >> N;

    A.resize(N);
    P.resize(N);
    tree.resize(N);
    ranges.resize(N);

    for (int i = 0; i < N; ++i) {
        cin >> P[i];
        if (i > 0) tree[P[i]].push_back(i);
    }

    for (int i = 0; i < N; ++i) {
        cin >> A[i];
    }

    set<int> values;
    dfs(0, -1, values);

    cout << result << endl;
    return 0;
}

path coverage range ✅

Colourd ball

Kth palindrome

String container

Play with stacks

Perfectly even

Divisible string

Kth Palindrome Code
Infosys SP Exam
C++


#include <iostream>
#include <vector>


using namespace std;

const int MOD = 1000000007;

int v(int k) {
    return (1 << k) - 1;
}


bool p(int n) {
    while (n % 2 == 0) n /= 2;
    return n == 1;
}


long long s(int n) {
    vector<long long> dp(n + 1, 0);
    dp[0] = 1;
   
    for (int i = 1; i <= n; ++i) {
        for (int k = 1; v(k) <= i; ++k) {
            int l = v(k - 1) - 1;
            if (i >= 2 * l + 1) {
                dp[i] = (dp[i] + dp[i - (2 * l + 1)] * k) % MOD;
            }
        }
    }
   
    return dp[n];
}


int main() {
    int n;
    cin >> n;
   
    cout << s(n) << endl;
   
    return 0;
}

Kth Palindrome Code
Infosys SP Exam
C++

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