Problem Description
Given a sorted array of distinct integers A and an integer B, find and return how many pair of integers ( A[i], A[j] ) such that i != j have sum equal to B.
Solution
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public class Solution {
public int solve(int[] A, int B) {
int count=0;
int i=0;
int j=A.length-1;
while(i<j){
int sum = A[i] + A[j];
if(sum == B){ //If we got the required sum, move both the pointers since there are distinct elements.
count++;
i++;
j--;
}else{
if(sum<B){//If currentSum is lesser, move index i to get a larger value
i++;
}else{
j--;
}
}
}
return count;
}
}