Tanuja VIn this problem, we would be applying the two binary search to check whether the target element is present or not. If present, we would be moving the pointers forward & backward to find the occurrence of the element (first & last).
class Solution {
public int[] searchRange(int[] nums, int target) {
int res[] = {-1, -1};
int n = nums.length;
int low = 0, high = n-1;
while(low<=high){
int mid = low + (high-low)/2;
if(nums[mid]==target){
res[0] = mid;
high = mid-1;
}
else if(nums[mid]<target)
low = mid+1;
else
high = mid-1;
}
low = 0;
high = n-1;
while(low<=high){
int mid = low + (high-low)/2;
if(nums[mid]==target){
res[1] = mid;
low = mid+1;
}
else if(nums[mid]<target)
low = mid+1;
else
high = mid-1;
}
return res;
}
}
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