Given array of n integers and given a number X, find all the unique pairs of elements (a,b), whose summation is equal to X.
The following is my solution, it is O(nLo
Here is Java code for three approaches:
1. Using Map O(n), HashSet can also be used here.
2. Sort array and then use BinarySearch to look for complement O(nLog(n))
3. Traditional BruteForce two loops O(n^2)
public class PairsEqualToSum {
public static void main(String[] args) {
int a[] = {1,10,5,8,2,12,6,4};
findPairs1(a,10);
findPairs2(a,10);
findPairs3(a,10);
}
//Method1 - O(N) use a Map to insert values as keys & check for number's complement in map
static void findPairs1(int[]a, int sum){
Map pairs = new HashMap();
for(int i=0; i0 && foundAtIndex != i) //to avoid situation where binarySearch would find the original and not the complement like "5"
System.out.println("("+a[i]+","+(sum-a[i])+")");
}
}
//Method 3 - Brute Force O(n^2)
static void findPairs3(int[]a, int sum){
for(int i=0; i