Saturday 14 October 2017

Java Code determines whether or not there exist two elements in S whose sum is exactly x

Problem:

For a given a set S of n integers and another integer x, determines whether or not there exist two elements in S whose sum is exactly x.

Solution:

package com.basics;

import java.util.Scanner;

public class FindExactSum {

static int[] array;
int arrayLimit;
int x;

public static void main(String[] args) {
FindExactSum findExactSum = new FindExactSum();
findExactSum.initializeTheArrays();
System.out.println("is the array contains two elements whose sum is exactly equal to given x?? " + findExactSum.isExactSumAvaiableInTheArray());
}

public void initializeTheArrays(){
Scanner scanner = new Scanner(System.in);
System.out.println("enter the array limit");
arrayLimit = scanner.nextInt();
array = new int[arrayLimit];
System.out.println("enter the array elements");
for (int i = 0; i < array.length; i++) {
System.out.println("enter the element");
array[i] = scanner.nextInt();
}

System.out.println("enter the x value");
x = scanner.nextInt();
scanner.close();
}

public boolean isExactSumAvaiableInTheArray(){
boolean flag = false;
for (int i = 0; i < array.length; i++) {
for (int j = i; j < array.length; j++) {
if(array[i] + array[j] == x){
flag = true;
break;
}
}
}
return flag;
}

}


Output:

enter the array limit
3
enter the array elements
enter the element
12
enter the element
12
enter the element
3
enter the x value
14
is the array contains two elements whose sum is exactly equal to given x?? false


enter the array limit
4
enter the array elements
enter the element
12
enter the element
33
enter the element
2
enter the element
33
enter the x value
35
is the array contains two elements whose sum is exactly equal to given x?? true

1 comment:

  1. time Complexity is O(N^2) where N is length of array. This can be improve as below :

    Map map = new HashMap<>();

    public boolean isExactSumAvaiable(int[] array, int sum) {
    int n = array.length;
    int result = 0;
    int value = 0;
    for(int i = 0; i < n; i++) {
    value = array[i];
    result = sum - value;
    if(map.get(result) != null) {
    return true;
    } else {
    /** map.put(result, map.get(result) == null ? 1 : map.get(result) +1); */
    //I am hard coding the value in map
    map.put(result, 1);
    }

    }
    return false;
    }

    Time Complexity is O(N) and Space Complexity is O(N)

    ReplyDelete

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