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import math
from collections import Counter
def solve(n, chars):
    if n == 1:
        return 1
    ans = Counter(chars)
    a = [freq for freq in ans.values() if freq > 0]
    if len(a) == 1:
        return a[0]
    sum_gcd = 0
    len_freqs = len(a)
    for i in range(len_freqs):
        for j in range(i + 1, len_freqs):
            sum_gcd += math.gcd(a[i], a[j])
    return sum_gcd

Phone pe โœ…
def ok(n, arr):
    acha = sum(arr)
    theekhai = 0
    for i in range(n):
        acha = acha - theekhai - arr[i]
        if theekhai == acha:
            return i
        theekhai += arr[i]
    return -1

Phonepe โœ…
def solve(n, plants, k):
    uff = 0
    a = k
    for i in range(n):
        if plants[i] <= a:
            a -= plants[i]
            uff += 1
        else:
            steps += 2 * i + 1
            a = k - plants[i]
    return uff

Phone pe โœ…
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Subject Format of mail : Name | Company | Experience | Preferred Location | SDE l / SDE ll

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#include <iostream>
#include <vector>
#include <queue>
#include <climits>
using namespace std;
const vector<pair<int, int>> directions = {{0, -1}, {0, 1}, {-1, 0}, {1, 0}};
const vector<char> dirChar = {'L', 'R', 'U', 'D'};
int minCostToReachDestination(int N, vector<vector<char>>& Dir, int Sx, int Sy, int Dx, int Dy) {
    priority_queue<pair<int, pair<int, int>>, vector<pair<int, pair<int, int>>>, greater<>> pq;
    vector<vector<int>> cost(N, vector<int>(N, INT_MAX));
    pq.push({0, {Sx, Sy}});
    cost[Sx][Sy] = 0;
   
    while (!pq.empty()) {
        auto [curCost, pos] = pq.top();
        pq.pop();
       
        int x = pos.first;
        int y = pos.second;
                if (x == Dx && y == Dy) {
            return curCost;
        }
            for (int i = 0; i < 4; ++i) {
            int nx = x + directions[i].first;
            int ny = y + directions[i].second;
           
            if (nx >= 0 && nx < N && ny >= 0 && ny < N) {
                int newCost = curCost + (Dir[x][y] == dirChar[i] ? 0 : 1);
                if (newCost < cost[nx][ny]) {
                    cost[nx][ny] = newCost;
                    pq.push({newCost, {nx, ny}});
                }
            }
        }
    }
   
    return -1;
}

Cost optimization
Google โœ…
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