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Find First and Last Position of Element in Sorted Array with Explanation | Java

Find First and Last Index Occurrence of given target from Java Array using Binary Search

Find First and Last Index Occurrence of given target from Java Array using Binary Search

Problem Description :

Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value.

If target is not found in the array, return [-1, -1].

Example 1 :

Input: nums = [5, 7, 7, 8, 8, 8, 10], target = 8
Output: [3, 5]

Example 2 :

Input: nums = [5, 7, 7, 8, 8, 8, 10], target = 6
Output: [-1, -1]

We can solve this problem using brute force approach but it will take more time than binary search. So lets see binary search approach.

Solution 1 : Finding First and Last Position of Element in Array.


import java.util.Scanner;

public class FindFirstAndLastPosition {

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);

        System.out.println("Enter array size : ");
        int size = sc.nextInt();

        int[] array = new int[size];

        // Taking user inputs
        System.out.println("Enter array elements :");
        for (int i = 0; i < array.length; i++) {
            array[i] = sc.nextInt();
        }

        System.out.println("Enter target : ");
        int target = sc.nextInt();

        int[] result = new int[2];
       
        result[0] = searchFirstElement(array, target);
        result[1] = searchLastElement(array, target);
       
        System.out.println("["+result[0]+", "+result[1]+"]");

    }

    private static int searchFirstElement(int[] nums, int target) {
       
        int index = -1;
       
        if (nums.length == 0) {
            return index;
        }

        int start = 0;
        int end = nums.length-1;

        while (start <= end) {
            int mid = (end + start) / 2;
           
            if (nums[mid] >= target) {
                end = mid - 1;
            } else {
                start = mid + 1;
            }
           
            if (nums[mid] == target) {
                index = mid;
            }
        }

        return index;
    }

    private static int searchLastElement(int[] nums, int target) {

        int index = -1;
       
        if (nums.length == 0) {
            return index;
        }

        int start = 0;
        int end = nums.length-1;

        while (start <= end) {
            int mid = (end + start) / 2;
           
            if (nums[mid] <= target) {
                start = mid + 1;
            } else {
                end = mid - 1;
            }
           
            if (nums[mid] == target) {
                index = mid;
            }
        }
        return index;
    }
}

Output :

Enter array size :
6
Enter array elements :
1 2 3 8 8 10
Enter target :
8
[3 , 4]

------------------------------

Enter array size :
10
Enter array elements :
1 2 3 5 6 8 10 11 18 20
Enter target :
4
[-1, -1]

Solution Explanation :

  • Here we are using two methods, searchFirstElement for finding first and searchLastElement for finding last elements in given array. In both method, Binary search is used.
  • In searchFirstElement method,
    • We have to search till first element until current array element < target for finding first target.
  • In searchLastElement method,
    • We have to search till last element until current array element > target for finding last given target.
  • At last return index from both methods.

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