# 1. Introduction

This blog post discusses how to efficiently find the starting and ending positions of a target value in a sorted array. This problem is a variant of binary search and requires an understanding of how to modify standard binary search to find boundary conditions.

## Problem

Given an array a sorted in ascending order, find the starting and ending position of a given target value. If the target is not found in the array, return [-1, -1]. The solution must have O(log n) runtime complexity.

Example 1:

```Input: a = [1, 4, 4, 10, 10, 15, 20], target = 10
Output: [3, 4]
```

Example 2:

```Input: a = [1, 4, 4, 10, 10, 15, 20], target = 15
Output: [5, 5]
```

Example 3:

```Input: a = [1, 4, 4, 10, 10, 15, 20], target = 0
Output: [-1, -1]
```

# 2. Solution Steps

1. Apply binary search to find the first occurrence of the target.

2. Apply binary search to find the last occurrence of the target.

3. Modify the binary search to adjust the search range based on the comparison with the target.

# 3. Code Program

``````public class Solution {

// Main method for testing
public static void main(String[] args) {
int[] a1 = {1, 4, 4, 10, 10, 15, 20};
System.out.println("First and last position of 10: " + Arrays.toString(searchRange(a1, 10)));

int[] a2 = {1, 4, 4, 10, 10, 15, 20};
System.out.println("First and last position of 15: " + Arrays.toString(searchRange(a2, 15)));

int[] a3 = {1, 4, 4, 10, 10, 15, 20};
System.out.println("First and last position of 0: " + Arrays.toString(searchRange(a3, 0)));
}

// Method to find the starting and ending position of a target value
public static int[] searchRange(int[] a, int target) {
int[] result = new int[]{-1, -1};
result[0] = findFirst(a, target);
result[1] = findLast(a, target);
return result;
}

// Helper method to find the first occurrence of target
private static int findFirst(int[] a, int target) {
int idx = -1;
int start = 0, end = a.length - 1;
while (start <= end) {
int mid = start + (end - start) / 2;
if (a[mid] >= target) {
end = mid - 1;
} else {
start = mid + 1;
}
if (a[mid] == target) idx = mid;
}
return idx;
}

// Helper method to find the last occurrence of target
private static int findLast(int[] a, int target) {
int idx = -1;
int start = 0, end = a.length - 1;
while (start <= end) {
int mid = start + (end - start) / 2;
if (a[mid] <= target) {
start = mid + 1;
} else {
end = mid - 1;
}
if (a[mid] == target) idx = mid;
}
return idx;
}
}
``````

### Output:

```First and last position of 10: [3, 4]
First and last position of 15: [5, 5]
First and last position of 0: [-1, -1]
```

### Explanation:

The searchRange method uses two binary searches: findFirst to locate the first occurrence of the target, and findLast to find the last occurrence.

For the array [1, 4, 4, 10, 10, 15, 20], the target 10 is found at indices 3 and 4, thus returning [3, 4].

For target 15, the method returns [5, 5], and for a non-existent target 0, it returns [-1, -1].

This approach efficiently solves the problem with O(log n) complexity.