Problem Description
Given N distinct sorted elements, check if there exists a pair (i,j) such that A[i] - A[j] = k and k>0. (i!=j)
Solution
Two important thing in question is to figure out where to initilize the two pointers.
If we keep i=0, j=n-1:
diff = A[j] - A[i]. Let’s say that the diff is smaller than k. If we move i to right, the difference will be smaller. If we move j to the left, we have the same problem. So, we cannot make a decision using this appraoch. To address this problem, we can initialize the i and j to 0 and 1 respectively.
If k is negative, we can simply take mod(k) and do the same thing.
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package com.gauk;
public class TwoPairDifference {
public static void main(String[] args) {
int[] a = new int[]{-3,0,1,3,6,8,11,14,18,25};
int[] b = new int[]{1,4,6};
System.out.println(new TwoPairDifference().check(a, 5));
System.out.println(new TwoPairDifference().check(b, 2));
}
//Given N distinct sorted elements, check if there exists a pair (i,j) such that A[i] - A[j] = k and k>0. (i!=j)
public boolean check(int[] arr, int k){
int i=0;
int j=1;
while(j<arr.length){
int diff = arr[j] - arr[i];
if(diff == k){
System.out.println(i + " " + j);
return true;
}
else if(diff < k){
j++;
}else{
i++;
}
if(i==j){
j++;
}
}
return false;
}
}