samarthsewlani

n = int(input())
lst = []
for i in range(n):
    lst.append([int(x) for x in input().split()])
a, b = 0, 0
for i in range(n):
    a += lst[i][i]
    b += lst[i][n-i-1]
print(abs(a-b))

samarthsewlani

import java.util.*;

public class Solution {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int matrix[][] = new int[n][n];
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                matrix[i][j] = sc.nextInt();

samarthsewlani

n = 3
alice = [int(x) for x in input().split()]
bob = [int(x) for x in input().split()]
aliceScore, bobScore = 0, 0
for i in range(n):
    a, b = alice[i], bob[i]
    if(a>b):
        aliceScore+=1
    elif(b>a):
        bobScore+=1

samarthsewlani

// https://www.hackerrank.com/contests/cp-24-25-batch-coding-fundamentals-classwork/challenges/compare-the-triplets/problem
import java.util.*;

public class Solution {
    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);
        // Step 1: Take inputs
        int n = 3;
        int alice[] = new int[n], bob[] = new int[n];
        for(int i=0;i<n;i++)    alice[i] = sc.nextInt();

samarthsewlani

//https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/
class Solution {
    public int numPairsDivisibleBy60(int[] time) {
        int n = time.length, count = 0;
        for(int i=0;i<n;i++)       time[i] %= 60;
        Map<Integer, Integer> visited = new HashMap<>();
        for(int i=0;i<n;i++){
            int value = (60-time[i])%60;
            if(visited.containsKey(value)){
                count += visited.get(value);

samarthsewlani

//https://leetcode.com/problems/longest-palindrome-by-concatenating-two-letter-words/
class Solution {
    public int longestPalindrome(String[] words) {
        int n = words.length, count = 0;
        Map<String, Integer> visited = new HashMap<>();
        for(int i=0;i<n;i++){
            String word = words[i];
            String reverseWord = word.charAt(1) + "" + word.charAt(0) ;
            if(visited.containsKey(reverseWord) && visited.get(reverseWord)>0){
                count += 4;

samarthsewlani

//https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/description/
class Solution {
    public String removeDuplicates(String s, int k) {
        Deque<Character> stack = new ArrayDeque<>();
        Deque<Integer> nums = new ArrayDeque<>();
        stack.push('#');        nums.push(1);
        int n = s.length();
        for(int i=0;i<n;i++){
            if(s.charAt(i) == stack.peek()){
                nums.push(nums.peek()+1);

samarthsewlani

import sys
INFINITY = sys.maxsize

def travelling_salesperson_problem(weight, n, i, unvisited_cities):
    if(len(unvisited_cities) == 0):
        # If no cities left to traverse return to starting city 1
        # Assumption we always start and end tour at city 1
        dp[i][tuple()] = weight[i-1][0]
        return weight[i-1][0]
    

samarthsewlani

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#define MAX 100

int INFINITY = INT_MAX;

int X[MAX];

int compare(const void* num1, const void* num2) // comparing function  

samarthsewlani

#include <stdio.h>
#include <limits.h>
#define MAX 100

int INFINITY = INT_MAX;

struct Edge{
    int u, v, weight;
};