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# LeetCode#34: Find First and Last Position of Element in Sorted Array

This LeetCode problem is a great practice problem to implement a binary search for a value in a sorted array with duplicate values. It asks to find the indices of the first and start the occurrence of a value in the sorted array with possibly duplicate values.

It can be solved in a naive manner to search for the start and end in a linear runtime O(n). But a better solution is to use Binary Search which runs in O(logn) time.

A typical binary search method looks like this if the array has unique and sorted numbers.

``````private int binarySearch(int[] nums, int target, int l, int r) {
int result = -1;
while (l<=r){
int mid = (l+r)/2;
if(nums[mid]==target){
return mid;
} else if (nums[mid] < target) {
l = mid + 1;
} else {
r = mid - 1;
}
}
return result;
}``````

But for this problem, the Binary Search requires a little modification because of the duplicates and also that we can design it in a way to look for not only the target value but also specifically search for the starting and ending occurrence. So check out the solution below.

``````public class SearchForARange {
public int[] searchRange(int[] nums, int target) {
int[] res = new int[]{-1,-1};
if (nums.length==0) return res;
res[0] = binarySearchLeft(nums, target, 0, nums.length-1);
if (res[0]>=0) {
res[1] = binarySearchRight(nums, target, res[0], nums.length-1);
}
return res;
}

private int binarySearchLeft(int[] nums, int target, int l, int r) {
int result = -1;
while(l<=r){
int mid = (l+r)/2;
if (nums[mid] == target){
result = mid;
r = mid-1;

} else if (nums[mid] > target){
r = mid - 1;
} else {
l = mid + 1;
}
}
return result;
}

private int binarySearchRight(int[] nums, int target, int l, int r) {
int result = -1;
while (l<=r){
int mid = (l+r)/2;
if(nums[mid]==target){
result = mid;
l = mid + 1;
} else if (nums[mid] > target) {
r = mid - 1;
} else {
l = mid + 1;
}
}
return result;
}
}``````

For understanding the logic, check out this post.