CareerCup

function ij(A, B) {
"use strict"
        memo = []
        for (i = 0; i < A.length; i++) {
                for (j = i+1; j < B.length; j++) {
                        if (A[i] > B[j]) {
                                memo.push("" + (i +1) + "," + (j +1))
                        }
                }
        }
        var memo
        , i
        , j
        return memo
}
console.log(ij([1, 2, 3, 4], [5, 6, 2, 1]))

import sys
a=[int(x) for x in sys.stdin.readline().split('')]
b=[int(x) for x in sys.stdin.readline().split('')]
l=[]
for i in range(0,len(a)):
	for j in range(i+1,len(b)):
		if a[i]>b[j]:
			l.append((a[i],b[j]));
print l

Solution in C#

public int[,] Pairs(int[] arrayOne, int[] arrayTwo)
        {
            List<int> list = new List<int>();
            for (int i = 0; i < arrayOne.Length; i++)
            {
                for (int j = i + 1; j < arrayTwo.Length; j++)
                {
                    if (arrayOne[i] > arrayTwo[j])
                    {
                        list.Add(arrayOne[i]);
                        list.Add(arrayTwo[j]);
                    }
                }
            }
            int listLenght = list.Count / 2;
            int[,] returnArray = new int[listLenght / 2, 2];
            for (int i = 0; i < listLenght; i += 2)
            {
                returnArray[i / 2, 0] = list[2 * i];
                returnArray[i / 2, 1] = list[(2 * i) + 1];
            }
            return returnArray;
        }

Solution in javascript.

{{
function solve(A, B) {
var ret = [];
for (var j = 0; j <= B.length; j++) {
for (var i = 0; (i < j && i < A.length); i++) {
if (A[i] < B[j]) {
ret.push({
i: i,
j: j
});
}
}
}
return ret;
}
}}

Solution in javascript.

{{
function solve(A, B) {
var ret = [];
for (var j = 0; j <= B.length; j++) {
for (var i = 0; (i < j && i < A.length); i++) {
if (A[i] < B[j]) {
ret.push({
i: i,
j: j
});
}
}
}
return ret;
}
}}

Solution in javascript.

function solve(A, B) { 
var ret = []; 
for (var j = 0; j <= B.length; j++) { 
for (var i = 0; (i < j && i < A.length); i++) { 
if (A[i] < B[j]) { 
ret.push({ 
i: i, 
j: j 
}); 
} 
} 
} 
return ret; 
}

Solution in java :

public class PairMatch {
	
	public static String [][] PairFind(int a[],int b[])
	{
		String c[][]=new String[a.length+b.length][a.length+b.length];
		
		for(int i=0;i<a.length;i++)
		{
			
			for(int j=i+1;j<b.length;j++)
			{
			
				if(a[i]>b[j])
				{
					c[i][j]=""+i+","+j;
				
				}
			}
		}
		return c;
	}

solution in java

public class Greater {
	
	public static void main(String[] args) {
		
		int a[]={10,02,30,51,45,89,12};
		int b[]={35,20,15,46,89,74,5};
		for (int i = 0; i < a.length; i++) {
			
			for (int j = 0; j < b.length; j++) {
				{
					if(a[i]>b[j]&&i<j)
					{
						System.out.println(i+" "+j);
						
						
					}
				}
				
			}
			
		}
	}

}

public static void main(String[] args) {
		
		int[] arr1 = { 1 ,6,4,2,6,3};
		int[] arr2 = {5,8,3,9,6,2};
		
		for(int i=0;i<arr1.length;i++) {
			
			for(int j = i+1;j<arr2.length;j++) {
				
				if(arr1[i]<arr2[j])
					System.out.println("("+i+","+j+")");
			}
		}
	}

all of the i*j solutions are nice but if you sort the lists storing (index, value) you can then run a binary search to find where in the first array the value is smaller.

Nice thing about this is that now the solutions can build on each other.

public HashMap<Integer, Integer> Compare(int A[], int B[])	{
	HashMap<Integer, Integer> pairs = new HashMap<Integer, Integer>();
	for(int i=0; i<A.length; i++)	{
		for(int j=1; j<B.length; j++)	{
			if(A[i]>B[j])	{
				pairs.put(A[i], A[j]);
			}
		}
	}
	return pairs;
}

#include<iostream>
using namespace std;
int main()
{
int a[] = {1,2,3,55,6,4,2,1,6,7,9};
int b[] = {1,4,22,5,6,7,8,9,0,5,3};
int n = sizeof(a)/sizeof(a[1]);
int i,j;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
if(i < j && a[i]>b[j])
cout<<"("<<i<<","<<j<<")"<<endl;
}
}

return 0;
}

using binary indexed tree would be the most efficient to solve this

using binary indexed tree would be the most efficient to solve this

using binary indexed tree would be the most efficient to solve this

public static ArrayList<String> count(int[] a, int[] b){
        ArrayList<String> list = new ArrayList<String>();
        for(int i=0; i<a.length; i++){
            for(int j=i+1; j<b.length; j++){
                if(a[i]>b[j])
                list.add(i+","+j);
            }
        }
        return list;
     }