This question is from geeksforgeeks :
Here’s the link: https://practice.geeksforgeeks.org/problems/find-all-pairs-whose-sum-is-x5808/1#
Q) Given two unsorted arrays A of size N and B of size M of distinct elements, the task is to find all pairs from both arrays whose sum is equal to X.
import java.util.*;
import java.lang.*;
import java.io.*;
class pair {
long first, second;
public pair(long first, long second)
{
this.first = first;
this.second = second;
}
}
class GFG {
public static void main(String[] args) throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine().trim());
while(t-->0)
{
StringTokenizer stt = new StringTokenizer(br.readLine());
long N = Long.parseLong(stt.nextToken());
long M = Long.parseLong(stt.nextToken());
long X = Long.parseLong(stt.nextToken());
long A[] = new long[(int)(N)];
long B[] = new long[(int)(M)];
String inputLine[] = br.readLine().trim().split(" ");
for (int i = 0; i < N; i++) {
A[i] = Long.parseLong(inputLine[i]);
}
String inputLine1[] = br.readLine().trim().split(" ");
for (int i = 0; i < M; i++) {
B[i] = Long.parseLong(inputLine1[i]);
}
Solution obj = new Solution();
pair [] answer = obj.allPairs(A, B, N, M, X);
int sz = answer.length;
if(sz==0)
System.out.println(-1);
else{
StringBuilder output = new StringBuilder();
for(int i=0;i<sz;i++){
if(i<sz-1)
output.append(answer[i].first +" "+ answer[i].second + ", ");
else
output.append(answer[i].first +" "+ answer[i].second);
}
System.out.println(output);
}
}
}
}
// } Driver Code Ends
//User function Template for Java
/*
class pair {
long first, second;
public pair(long first, long second)
{
this.first = first;
this.second = second;
}
}
*/
class Solution {
public pair[] allPairs( long A[], long B[], long N, long M, long X) {
//MY CODE STARTS FROM HERE
ArrayList<pair> list = new ArrayList<>();
HashSet<Long> set = new HashSet<>();
for(long i : A){
set.add(i);
}
for(int i=0;i<M;i++){
if(set.contains(X-B[i])){
list.add(new pair(X-B[i], B[i]));
}
}
pair arr[] = new pair[list.size()];
arr = list.toArray(arr);
for(int j=0;j<(arr.length-1);j++){
if(arr[j]>arr[j+1]){
pair temp = arr[j];
arr[j] = arr[j+1];
arr[j+1] = temp;
j -= 1;
}
}
return arr;
}
}
I have to sort the result in ascending order but the datatype for array here is user-defined (i.e. pair). So I am getting this error :
prog.java:100: error: bad operand types for binary operator ‘>’ if(arr[j]>arr[j+1]){ ^ first type: pair second type: pair 1 error
Please help me sort the array. Note: Time complexity should be O(nlogn)
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Answer
Why use the custom sort when you can use built-in sort in Java list.
Just use list.sort(Comparator.comparing(p -> ((pair) p).first).thenComparing(p -> ((pair) p).second));
to sort the pair list.
public pair[] allPairs( long A[], long B[], long N, long M, long X) {
// Your code goes here
//MY CODE STARTS FROM HERE
ArrayList<pair> list = new ArrayList<>();
HashSet<Long> set = new HashSet<>();
for(long i : A){
set.add(i);
}
for(int i=0;i<M;i++){
if(set.contains(X-B[i])){
list.add(new pair(X-B[i], B[i]));
}
}
list.sort(
Comparator
.comparing(p -> ((pair) p).first)
.thenComparing(p -> ((pair) p).second)
);
return list.toArray(pair[]::new);
}