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Update ThreeSumProblem
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‎ThreeSumProblem

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********************************** I. Using Naive Approach ********************************************
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/*
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* Question:
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Given an array of integers, return one set of 3 elements such that the 3 numbers add up to zero
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(VERY IMP NOTE: REPETITIONS OF ELEMENTS ARE ALLOWED)
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{10, -2, -1, 3} -> true
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{10, -2, 1} -> true -2 + 1 +1 =0
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*
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* Source: http://www.geeksforgeeks.org/find-a-triplet-that-sum-to-a-given-value/
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*
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* Algorithm : This is a THREE SUM PROBLEM:
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* Generate all possible three set sums and check whether any of them matches with the given sum.
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* 1. Run first loop from i to arraySize
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* 1.1 Run second loop from j=i+1 to arraySize
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* 1.1.1 Run third loop from k=j+1 to arraySize
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* Check if array[i]+array[j]+array[k]=Sum
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*
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*/
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package ThreeSumProblem;
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import java.util.Scanner;
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public class UsingNaiveApproach {
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public static void main(String[] args) {
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Scanner in = new Scanner(System.in);
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try{
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System.out.println("Enter the number of elements in the array");
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int n = in.nextInt();
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int[] array = new int[n];
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System.out.println("Enter the elements of the array");
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for(int i=0;i<n;i++)
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array[i]=in.nextInt();
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System.out.println("Enter the sum that you need to find in the array");
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int sum = in.nextInt();
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System.out.println("The elements whose addition returns the sum are: ");
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usingNaiveApproach(array,sum);
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}
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finally{
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in.close();
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}
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}
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private static void usingNaiveApproach(int[] array, int sum) {
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outerloop: // label the outer loop
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for(int i=0;i<array.length;i++)
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for(int j=i+1;j<array.length;j++)
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for(int k=j+1;k<array.length;k++)
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if(array[i]+array[j]+array[k]==sum){
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System.out.println(array[i]+" "+array[j]+" "+array[k]);
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break outerloop; // break the outer loop so that all internal loops are also broken
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}
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}
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}
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/*
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* Analysis:
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* Time Complexity = O(n^3)
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* Space Complexity = O(1)
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*/
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********************************** II. Using Sorting ***********************************************
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/*
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* Question:
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Given an array of integers, return one set of 3 elements such that the 3 numbers add up to zero
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(VERY IMP NOTE: REPETITIONS OF ELEMENTS ARE ALLOWED)
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{10, -2, -1, 3} -> true
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{10, -2, 1} -> true -2 + 1 +1 =0
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*
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* Source: http://www.geeksforgeeks.org/find-a-triplet-that-sum-to-a-given-value/
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*
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* Algorithm : This is a THREE SUM PROBLEM:
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* Sort the array and then apply the Two Sum Logic
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* 1. Sort the array
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* 2. Run a loop starting from i=0 to (arraySize-2)
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* 2.1 Take a fixed element as fixedElement = array[i]
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* 2.2 Find the other two elements from the SORTED ARRAY CORNERS
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* low = [i+1] // Next Element
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* high = [arraySize-1] // last element of the array
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* 2.2.1 while(low<high)
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* Check if(fixed+low+high == sum). If yes then print the elements and break
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* if(fixed+low+high < sum). Then low++
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* if(fixed+low+high > sum). Then high--
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*/
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package ThreeSumProblem;
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import java.util.Arrays;
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import java.util.Scanner;
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public class UsingSorting {
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public static void main(String[] args) {
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Scanner in = new Scanner(System.in);
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try{
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System.out.println("Enter the number of elements in the array");
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int n = in.nextInt();
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int[] array = new int[n];
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System.out.println("Enter the elements of the array");
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for(int i=0;i<n;i++)
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array[i]=in.nextInt();
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System.out.println("Enter the sum that you need to find in the array");
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int sum = in.nextInt();
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System.out.println("The elements whose addition returns the sum are: ");
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usingSorting(array,sum);
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}
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finally{
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in.close();
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}
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}
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private static void usingSorting(int[] array, int sum) {
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int low=0;
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int high=0;
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/* Sort the elements */
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Arrays.sort(array);
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/* Now fix the first element one by one and find the
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other two elements */
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outerloop:
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for(int i=0;i<(array.length-2);i++){
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// To find the other two elements, start two index variables
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// from two corners of the array and move them toward each
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// other
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low=i+1;
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high = array.length-1;
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while(low<high){
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if(array[i]+array[low]+array[high]==sum){
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System.out.println(array[i]+" "+array[low]+" "+array[high]);
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break outerloop;
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}
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if(array[i]+array[low]+array[high]<sum)
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low++;
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if(array[i]+array[low]+array[high]>sum)
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high--;
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}
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}
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}
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}
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/*
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* Analysis:
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* Time Complexity = O(n^2) due to nested for and while loop
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* Space Complexity = O(1)
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*/
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**************************** III Solving Three Sum Problem Using HashMap ****************************************
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VERY IMP NOTE: Time Complexity of Three Sum Problem using:
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1. HashMap = O(n^3)
@@ -8,7 +161,6 @@ we can SAFELY use HashSet.
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2. HashSet = O(n^2)
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HashSet is used when array elements are unique.
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**************************** I. Solving Three Sum Problem Using HashMap ****************************************
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/*
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* Question:
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Given an array of integers, return one set of 3 elements such that the 3 numbers add up to zero
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array[i]=in.nextInt();
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System.out.println("Enter the sum that you need to find in the array");
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int sum = in.nextInt();
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System.out.println("The elements whose additon returns the sum are: ");
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System.out.println("The elements whose addition returns the sum are: ");
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findThreeSumUsingHashMap(array,sum);
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}
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finally{
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// iterate through the array elements and check whether the third is in the HashMap
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outerloop: // using label in Java to break the outer loop
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for(int i=0;i<array.length;i++)
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for(int j=i;j<array.length;j++)
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if(map.containsValue(-(array[i]+array[j]))){ // containsValue time complexity is O(n)
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System.out.print(array[i]+" "+array[j]+" "+(-(array[i]+array[j])));
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for(int j=i+1;j<array.length;j++) // start from the next element
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if(map.containsValue(sum-(array[i]+array[j]))){ // containsValue time complexity is O(n)
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System.out.print(array[i]+" "+array[j]+" "+(sum-(array[i]+array[j])));
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break outerloop; // we can use labels to break directly the outer loop (which in turn breaks inner loop)
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} // Otherwise, we would have to implement some mechanism to break both the loops simultaneously
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* Space Complexity = O(n)
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*/
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**************************** I. Solving Three Sum Problem Using HashSet ****************************************
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**************************** IV. Solving Three Sum Problem Using HashSet ****************************************
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/*
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* Question:
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Given an array of integers, return one set of 3 elements such that the 3 numbers add up to zero
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array[i]=in.nextInt();
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System.out.println("Enter the sum that you need to find in the array");
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int sum = in.nextInt();
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System.out.println("The elements whose additon returns the sum are: ");
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System.out.println("The elements whose addition returns the sum are: ");
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findThreeSumUsingHashSet(array,sum);
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}
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finally{
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// iterate through the array elements and check whether the third is in the HashMap
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outerloop: // using label in Java to break the outer loop
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for(int i=0;i<array.length;i++)
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for(int j=i;j<array.length;j++)
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if(set.contains(-(array[i]+array[j]))){ // time complexity of contains method of HashSet is O(1)
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System.out.print(array[i]+" "+array[j]+" "+(-(array[i]+array[j])));
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for(int j=i+1;j<array.length;j++) // start from the next element
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if(set.contains(sum-(array[i]+array[j]))){ // time complexity of contains method of HashSet is O(1)
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System.out.print(array[i]+" "+array[j]+" "+(sum-(array[i]+array[j])));
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break outerloop; // we can use labels to break directly the outer loop (which in turn breaks inner loop)
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} // Otherwise, we would have to implement some mechanism to break both the loops simultaneously
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