In this post, We are going to solve HackerRank Divisible Sum Pairs Problem. Given an array of integers and a positive integer k, determine the number of (I, j) pairs where I < J and ar [I] + is divisible by k.
Example
ar = [ 1, 2, 3, 4, 5, 6]
k = 5
Three pairs meet the criteria: [ 1, 5 ], [ 2, 3 ], and [ 4, 6 ].
Function Description
Complete the divisibleSumPairs function in the editor below.
divisibleSumPairs has the following parameter(s):
- int n: the length of the array ar
- int ar[n]: an array of integers
- int k: the integer divisor
Returns
– int: the number of pairs
Input Format
The first line contains 2 space-separated integers, n, and k.
The second line contains n space-separated integers, each a value of arr[I].
Constraints
2 < n < 100
1 < k < 100
1 < arr[I] < 100
Sample Input
STDIN Function
----- --------
6 3 n = 6, k = 3
1 3 2 6 1 2 ar = [1, 3, 2, 6, 1, 2]
Sample Output
5
Divisible Sum Pairs C Solutions
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
int main(){
int n;
int k;
int count = 0;
scanf("%d %d",&n,&k);
int *arr = malloc(sizeof(int) * n);
for(int i = 0; i < n; i++){
scanf("%d",&arr[i]);
}
for (int i = 0; i < n; i++)
{
for (int j = i + 1; j < n; j++)
{
if ((arr[i] + arr[j]) % k == 0)
{
count++;
}
}
}
printf("%d", count);
return 0;
}Divisible Sum Pairs C++ Solution
// Created by Kaidul Islam on 18/4/16.
// Copyright © 2016 Kaidul Islam. All rights reserved.
//
//
#include <algorithm>
#include <array>
#include <bitset>
#include <cassert>
#include <cctype>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <iomanip>
#include <iostream>
#include <limits>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
using namespace std;
#define rep(i, n) for(__typeof(n) i = 0; i < (n); i++)
#define rrep(i, n) for(__typeof(n) i = (n) - 1; i >= 0; --i)
#define rep1(i, n) for(__typeof(n) i = 1; i <= (n); i++)
#define FOR(i, a, b) for(__typeof(b) i = (a); i <= (b); i++)
#define forstl(i, s) for (__typeof ((s).end ()) i = (s).begin (); i != (s).end (); ++i)
#define SIZE(v) ((int)v.size())
#define INF (1 << 30)
#define PI acos(-1.0)
#define bitcnt __builtin_popcount
#define pb push_back
#define ppb pop_back
#define all(x) x.begin(), x.end()
#define mem(x, y) memset(x, y, sizeof x)
#define eps 1e-9
#define pii pair<int, int>
#define couple make_pair
#define vi vector<int>
#define vpii vector< pii >
#define si set<int>
#define SDi(x) sf("%d", &x)
#define SD2(x, y) sf("%d %d", &x, &y)
#define SD3(x, y, z) sf("%d %d %d", &x, &y, &z)
#define SDs(x) sf("%s", x)
#define pf printf
#define print(x) pf("%d ", x)
#define println(x) pf("%d\n", x)
#define output(x, y); pf("Case %d: %d", ++x, y)
#define newLine pf("\n")
#define sf scanf
#define READ(f) freopen(f, "r", stdin)
#define WRITE(f) freopen(f, "w", stdout)
#if ( _WIN32 or __WIN32__ )
#define LLD "%I64d"
#else
#define LLD "%lld"
#endif
#define SDl(x) sf(LLD, &x)
#define MAX5 100000
#define MAX7 10000000
#define MAX9 1000000000
#define MOD7 (MAX7 + 7)
#define MOD9 (MAX9 + 9)
typedef long long i64;
typedef unsigned long long ui64;
const i64 INF64 = (i64) 1E18;
template<typename T> string toStr(T n) { ostringstream oss; oss << n; oss.flush(); return oss.str(); }
template<typename T> T toInt(string s) { T n = 0; istringstream sin(s); sin >> n; return n; }
class TimeTracker {
clock_t start, end;
public:
TimeTracker() {
start = clock();
}
~TimeTracker() {
end = clock();
fprintf(stderr, "%.3lf s\n", (double)(end - start) / CLOCKS_PER_SEC);
}
};
struct Point {
int x, y;
Point(): x(0), y(0) {}
Point(int a, int b): x(a), y(b) {}
bool operator < (const Point& other) const {
return x < other.x;
}
};
// BitMask
int Set(int N, int pos) {
return N = N | (1 << pos);
}
int Reset(int N, int pos) {
return N = N & ~(1 << pos);
}
int Check(int N, int pos) {
return (N & (1 << pos));
}
int toggle(int N, int pos) {
if( Check(N, pos) )
return N = Reset(N, pos);
return N = Set(N, pos);
}
// direction array
//int dx[] = {0, -1, 0, 1};
//int dy[] = {-1, 0, 1, 0};
//int Dx[] = {0, -1, -1, -1, 0, 1, 1, 1};
//int Dy[] = {-1, -1, 0, 1, 1, 1, 0, -1};
//int row, col;
//inline bool isValid(int i, int j) {
// return i >= 0 and j >= 0 and i < row and j < col;
//}
/** Implementation **/
#define MAX 105
int arr[MAX];
int main() {
int n, k;
SD2(n, k);
rep(i, n) SDi(arr[i]);
int result = 0;
FOR(i, 0, n - 1) {
FOR(j, i + 1, n - 1) {
int sum = arr[i] + arr[j];
result += (sum % k == 0);
}
}
println(result);
return 0;
}Divisible Sum Pairs C Sharp Solution
using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
class Solution {
static void Main(String[] args) {
string[] tokens_n = Console.ReadLine().Split(' ');
int n = Convert.ToInt32(tokens_n[0]);
int k = Convert.ToInt32(tokens_n[1]);
string[] a_temp = Console.ReadLine().Split(' ');
int[] a = Array.ConvertAll(a_temp,Int32.Parse);
int pair = 0;
for(int i = 0; i<n; i++){
for(int j = i+1; j < n; j++){
if(j>i){
if((a[i] + a[j])%k == 0)
pair++;
}
}
}
Console.WriteLine(pair);
}
}Divisible Sum Pairs Java Solution
import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;
class Result {
/*
* Complete the 'divisibleSumPairs' function below.
*
* The function is expected to return an INTEGER.
* The function accepts following parameters:
* 1. INTEGER n
* 2. INTEGER k
* 3. INTEGER_ARRAY ar
*/
public static int divisibleSumPairs(int n, int k, List<Integer> ar) {
// Write your code here
int pairsCount = 0;
for(int i = 0; i < n; i++)
{
for(int j = i + 1; j < n; j++)
{
if((ar.get(i) + ar.get(j)) % k == 0)
{
pairsCount++;
}
}
}
return pairsCount;
}
}
public class Solution {
public static void main(String[] args) throws IOException {
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));
String[] firstMultipleInput = bufferedReader.readLine().replaceAll("\\s+$", "").split(" ");
int n = Integer.parseInt(firstMultipleInput[0]);
int k = Integer.parseInt(firstMultipleInput[1]);
List<Integer> ar = Stream.of(bufferedReader.readLine().replaceAll("\\s+$", "").split(" "))
.map(Integer::parseInt)
.collect(toList());
int result = Result.divisibleSumPairs(n, k, ar);
bufferedWriter.write(String.valueOf(result));
bufferedWriter.newLine();
bufferedReader.close();
bufferedWriter.close();
}
}Divisible Sum Pairs JavaScript Solution
process.stdin.resume();
process.stdin.setEncoding('ascii');
var input_stdin = "";
var input_stdin_array = "";
var input_currentline = 0;
process.stdin.on('data', function (data) {
input_stdin += data;
});
process.stdin.on('end', function () {
input_stdin_array = input_stdin.split("\n");
main();
});
function readLine() {
return input_stdin_array[input_currentline++];
}
/////////////// ignore above this line ////////////////////
const evenlyDivides = (x, y) => Number.isInteger(y / x);
const sum = (x, y) => x + y;
const getDivisibleSumPairs = (baseNum, k, arr) => {
return arr.filter((val) => evenlyDivides(k, sum(baseNum, val)));
};
const getDivisibleSumPairCount = (arr, k) => {
return arr.reduce((prev, curr, idx) => {
return prev + getDivisibleSumPairs(curr, k, arr.slice(idx + 1)).length;
}, 0);
};
function main() {
var n_temp = readLine().split(' ');
var n = parseInt(n_temp[0]);
var k = parseInt(n_temp[1]);
a = readLine().split(' ');
a = a.map(Number);
console.log(getDivisibleSumPairCount(a, k));
}Divisible Sum Pairs Python Solution
#! /usr/bin/python3
# https://www.hackerrank.com/contests/w20/challenges/divisible-sum-pairs
n, k = map(int, input().strip().split())
a = [int(i) for i in input().strip().split()]
ret = 0
for i in range(0, n):
for j in range(i + 1, n):
if (a[i] + a[j]) % k == 0:
ret = ret + 1
print(ret)Other Solutions
