#include <iostream>
#include <string>
#include <set>
#include <algorithm>
using namespace std;
string LexicographicOrder(string S) {
string result = "";
while (!S.empty()) {
set<char> seen;
string uniqueChars = "";
for (char c : S) {
if (seen.find(c) == seen.end()) {
uniqueChars += c;
seen.insert(c);
            }
        }
sort(uniqueChars.begin(), uniqueChars.end());
result += uniqueChars;
for (char c : uniqueChars) {
size_t pos = S.find(c);
if (pos != string::npos) {
S.erase(pos, 1);
            }
        }
    }
    return result;
}

Tiger Analytics ✅

#include <bits/stdc++.h>
using namespace std;
long long comb(int n, int r) {
    if (r > n) return 0;
    if (r == 0 || r == n) return 1;
    long long res = 1;
    for (int i = 0; i < r; i++) {
        res = res * (n - i) / (i + 1);
    }
    return res;
}
long long Balanced(int A, int B, int C) {
    long long s = 0;
    for (int p = 4; p <= min(A, C - 1); p++) {
    int w = C - p; 
    if (w >= 1 && w <= B) { 
    s += comb(A, p) * comb(B, w);
        }
    }
    return s;
} // Balance Mixture

✅

#include <iostream>
#include <string>
#include <cmath>
using namespace std;
int gap(char a, char b) {
    return abs(a - b);
}
string longestKInterspaceSubstring( string &word, int k) {
    string temp = "", maxSubstring = "";
    for (size_t i = 0; i < word.length(); i++) {
    temp += word[i];
    if (i < word.length() - 1 && gap(word[i], word[i + 1]) > k) {
    if (temp.length() > maxSubstring.length()) {
    maxSubstring = temp;
    }
    temp = "";
        }
    }
    if (temp.length() > maxSubstring.length()) {
        maxSubstring = temp;
    }

    return maxSubstring;
}

Longest K intersapace substring ✅
Paypal

public static int getMinProcessingTime(int[] data, int processTimeA, int processTimeB) {
        int n = data.length;
        int tds = 0;
        for (int i : data) {
            tds += i;
        }

        int[] dp = new int[tds + 1];
        for (int i = 0; i <= tds; i++) {
            dp[i] = Integer.MAX_VALUE;
        }
        dp[0] = 0;

        for (int size : data) {
            for (int j = tds - size; j >= 0; j--) {
                if (dp[j] != Integer.MAX_VALUE) {
                    dp[j + size] = Math.min(dp[j + size], dp[j] + size * processTimeA);
                }
            }
        }

        int minTime = Integer.MAX_VALUE;
        for (int sumA = 0; sumA <= tds; sumA++) {
            if (dp[sumA] != Integer.MAX_VALUE) {
                int sumB = tds - sumA;
                int timeB = sumB * processTimeB;
                minTime = Math.min(minTime, Math.max(dp[sumA], timeB));
            }
        }

        return minTime;
    }

Minimum Processing Time ✅

Swiggy is hiring for Associate Software Engineer.

If interested, fill the form: https://docs.google.com/forms/d/e/1FAIpQLSfz_6J2t7ES3KjCMex-_0_c12K74_ggpjGttFeuD8J5oZ80TA/viewform

from collections import Counter
def solve(n, s, k):
    freq = Counter(s)
    distinct_count = len(freq)
    if k > n:
        return -1
    if k == distinct_count:
        return 0
    operations = 0
    if k > distinct_count:
        operations = k - distinct_count
    else:
        extra = distinct_count - k
        freq_values = sorted(freq.values())
        operations = sum(freq_values[:extra])
    return operations

Distinct Character transformation ✅

import math
def solve(l, r):
    count = 0
    for x in range(l, r + 1):
        sqrt_floor = math.isqrt(x)
        if x % sqrt_floor == 0:
            count += 1
    return count

Lucky number ✅

Shellovation hiring Mobile / App Development Intern

2026/2027 batch passouts
38 k per month stipend

Apply Here : https://unstop.com/internships/mobile-app-developer-shellovation-private-limited-1270510?lb=riTZ96J&utm_medium=Share&utm_source=shortUrl

Marsh McLennan is hiring (only with referral)

Take referral on LinkedIn and apply!!

Company : Uber
Program: Uber She++ and UberSTAR Internship
Eligibility : 2027

Apply: https://flows.beamery.com/uber/apac-uber-she---2025-s_29svpsn

Complete details about eligibility

📌PayU is hiring!
Data Engineering Intern

Location: Mumbai, Bengaluru, Gurgaon
Qualification:
- A degree in Engineering, Statistics, Computer Science, Mathematics or other similar quantitative field from a premier institute
- Students in the 2nd / 3rd semester & final year of their Graduation/Post Graduation can apply for an internship

💻Apply Link: https://jobs.eu.lever.co/payu/ee2691f3-64c7-4251-9284-e91aa40b2932/

def ok(mid, N, M):
    count = 0
    for i in range(1, N + 1):
        count += min(M, mid // i)
    return count
def solve(N, M, K):
    low, high = 1, N * M
    while low < high:
        mid = (low + high) // 2
        if ok(mid, N, M) < K:
            low = mid + 1
        else:
            high = mid
    return low
N, M, K = map(int, input().split())
print(solve(N, M, K))

A lot to merge ✅
Media. Net

#include <iostream>
#include <vector>
using namespace std;
const int MOD = 1000000007;

struct node {
    int val;
    int lazy;
    node *left;
    node *right;

    node() {
        left = nullptr;
        right = nullptr;
        lazy = 0;
        val = 0;
    }
};

void build(node *&root, int s, int e) {
    root = new node();
    if (s == e) {
        root->val = 0;
    } else {
        int mid = (s + e) / 2;
        build(root->left, s, mid);
        build(root->right, mid + 1, e);
        root->val = root->left->val + root->right->val;
    }
}

void applyLazy(node* root, int s, int e) {
    if (root->lazy) {
        root->val = (e - s + 1) - root->val;
        if (s != e) {
            if (!root->left) root->left = new node();
            if (!root->right) root->right = new node();
            root->left->lazy ^= 1;
            root->right->lazy ^= 1;
        }
        root->lazy = 0;
    }
}

void update_range(node *root, int s, int e, int l, int r) {
    applyLazy(root, s, e);
    if (s > r || e < l) return;
    if (s >= l && e <= r) {
        root->lazy ^= 1;
        applyLazy(root, s, e);
        return;
    }
    int mid = (s + e) / 2;
    update_range(root->left, s, mid, l, r);
    update_range(root->right, mid + 1, e, l, r);
    root->val = root->left->val + root->right->val;
}

int query(node *root, int s, int e, int l, int r) {
    applyLazy(root, s, e);
    if (s > r || e < l) return 0;
    if (s >= l && e <= r) {
        return root->val;
    }
    int mid = (s + e) / 2;
    int leftQuery = query(root->left, s, mid, l, r);
    int rightQuery = query(root->right, mid + 1, e, l, r);
    return leftQuery + rightQuery;
}

int solve(int N, vector<vector<int>>& B) {
    node* root = nullptr;
    build(root, 0, N - 1);
    
    int sumQueries = 0;
    for (const auto& op : B) {
        int type = op[0];
        int l = op[1] - 1;
        int r = op[2] - 1;

        if (type == 0) {
            update_range(root, 0, N - 1, l, r);
        } else {
            sumQueries += query(root, 0, N - 1, l, r);
            sumQueries %= MOD;
        }
    }
    return sumQueries;
}

int main() {
    int N, Q;
    cin >> N >> Q;
    vector<vector<int>> B(Q, vector<int>(3));
    
    for (int i = 0; i < Q; ++i) {
        cin >> B[i][0] >> B[i][1] >> B[i][2];
    }

    int result = solve(N, B);
    cout << result << endl;

    return 0;
}

Mask updates
Media. Net✅

#include <bits/stdc++.h>
using namespace std;
void solve() {
    int T;
    cin >> T;
    while (T--) {
        int n, x;
        cin >> n >> x;
        vector<int> a(n);
        for (int i = 0; i < n; i++) {
            cin >> a[i];
        }
        vector<int> remainderCount(x, 0);
        for (int i = 0; i < n; i++) {
            remainderCount[a[i] % x]++;
        }
        int pairs = 0;
        pairs += remainderCount[0] / 2;
        for (int r = 1; r <= x / 2; r++) {
            if (r == x - r) { 
                pairs += remainderCount[r] / 2;
            } else {
                int minCount = min(remainderCount[r], remainderCount[x - r]);
                pairs += minCount;
                remainderCount[r] -= minCount;
                remainderCount[x - r] -= minCount;
            }
        }
        cout << pairs * 2 << endl;
    }
}

int main() {
    solve();
    return 0;
}

christmas celebration ✅