Hello coders, today we are going to solve Diagonal Difference HackerRank Solution which is a Part of HackerRank Algorithms Series.
Task
Given a square matrix, calculate the absolute difference between the sums of its diagonals.
For example, the square matrix arr is shown below:
1 2 3
4 5 6
9 8 9
The left-to-right diagonal = 1 + 5 +9 = 15. The right to left diagonal = 3 + 5 + 9 = 17. Their absolute difference is |15 – 17| = 2.
Function description
Complete the DiagonalDifference function in the editor below.
diagonalDifference takes the following parameter:
- int arr[n][m]: an array of integers
Return
- int: the absolute diagonal difference
Input Format
The first line contains a single integer, n, the number of rows and columns in the square matrix arr.
Each of the next n lines describes a row, arr[i], and consists of n space-separated integers arr[i][j].
Constraints
- -100 <= arr[i][j] <= 100
Output Format
Return the absolute difference between the sums of the matrix’s two diagonals as a single integer.
Sample Input
3
11 2 4
4 5 6
10 8 -12
Sample Output
15
Explanation
The primary diagonal is:
11
5
-12
Sum across the primary diagonal: 11 + 5 – 12 = 4
The secondary diagonal is:
4
5
10
Sum across the secondary diagonal: 4 + 5 + 10 = 19
Difference: |4 – 19| = 15
Note: |x| is the absolute value of x
Solution – Diagonal Difference Solution
C++
#include <bits/stdc++.h> using namespace std; string ltrim(const string &); string rtrim(const string &); vector<string> split(const string &); /* * Complete the 'diagonalDifference' function below. * * The function is expected to return an INTEGER. * The function accepts 2D_INTEGER_ARRAY arr as parameter. */ int diagonalDifference(vector<vector<int>> arr) { int s1 = 0; int s2 = 0; int n = arr.size(); for(int i=0;i<n;i++) { s1 += arr[i][i]; s2 += arr[i][n-i-1]; } return abs(s1 - s2); } int main() { ofstream fout(getenv("OUTPUT_PATH")); string n_temp; getline(cin, n_temp); int n = stoi(ltrim(rtrim(n_temp))); vector<vector<int>> arr(n); for (int i = 0; i < n; i++) { arr[i].resize(n); string arr_row_temp_temp; getline(cin, arr_row_temp_temp); vector<string> arr_row_temp = split(rtrim(arr_row_temp_temp)); for (int j = 0; j < n; j++) { int arr_row_item = stoi(arr_row_temp[j]); arr[i][j] = arr_row_item; } } int result = diagonalDifference(arr); fout << result << "\n"; fout.close(); return 0; } string ltrim(const string &str) { string s(str); s.erase( s.begin(), find_if(s.begin(), s.end(), not1(ptr_fun<int, int>(isspace))) ); return s; } string rtrim(const string &str) { string s(str); s.erase( find_if(s.rbegin(), s.rend(), not1(ptr_fun<int, int>(isspace))).base(), s.end() ); return s; } vector<string> split(const string &str) { vector<string> tokens; string::size_type start = 0; string::size_type end = 0; while ((end = str.find(" ", start)) != string::npos) { tokens.push_back(str.substr(start, end - start)); start = end + 1; } tokens.push_back(str.substr(start)); return tokens; }
Python
#!/bin/python3 import math import os import random import re import sys # # Complete the 'diagonalDifference' function below. # # The function is expected to return an INTEGER. # The function accepts 2D_INTEGER_ARRAY arr as parameter. # def diagonalDifference(arr): # Write your code here d1 = sum([arr[x][x] for x in range(len(arr))]) d2 = sum([arr[x][n - 1 - x] for x in range(len(arr))]) return(abs(d1 - d2)) if __name__ == '__main__': fptr = open(os.environ['OUTPUT_PATH'], 'w') n = int(input().strip()) arr = [] for _ in range(n): arr.append(list(map(int, input().rstrip().split()))) result = diagonalDifference(arr) fptr.write(str(result) + '\n') fptr.close()
Disclaimer: The above Problem (Diagonal Difference) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.
I found a way that only needs to loop through the matrix once.
def diagonalDifference(arr):
difference = 0
for index, cont in enumerate(arr):
difference += cont[index] – cont[-index – 1]
return abs(difference)