Question | 34. Find First and Last Position of Element in Sorted Array – Leetcode Javascript Solution |
Difficulty | Medium |
Question Type | Array |
Question Link | https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/ |
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].
You must write an algorithm with O(log n) runtime complexity.
Example 1:
Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]
Example 2:
Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]
Example 3:
Input: nums = [], target = 0 Output: [-1,-1]
Constraints:
0 <= nums.length <= 105
-109 <= nums[i] <= 109
nums is a non-decreasing array.
-109 <= target <= 109
Solution 1 (Using Native Array Sort) :
var searchRange = function(nums, target) {
let found = [];
for(let i=0; i<nums.length; i++){
if(nums[i] === target){
found.push(i);
}
}
if(found.length === 1) return [found[0], found[0]];
const start = found.length > 1 ? found[0] : -1;
const end = found.length > 1 ? found[found.length-1] : -1 ;
return [start, end];
};
Complexity Analysis :
Time complexity: O(n) Space complexity: O(n)
Solution 2 ( Binary Search ) ==> Optimised :
// Binary search for searching
var searchRange = function (nums, target) {
if (nums.length < 1) {
return [-1, -1];
}
let firstPos = binarySearch(nums, 0, nums.length - 1, target);
let start = firstPos, end = firstPos, tmp1, tmp2;
if (start === -1) {
return [-1, -1];
}
while (start !== -1) {
tmp1 = start;
start = binarySearch(nums, 0, start - 1, target);
}
start = tmp1;
while (end !== -1) {
tmp2 = end;
end = binarySearch(nums, end + 1, nums.length - 1, target);
}
end = tmp2;
return [start, end];
};
const binarySearch = function (nums, left, right, target) {
while (left <= right) {
const mid = Math.floor((left + right) / 2);
if (target === nums[mid]) {
return mid;
} else if (nums[mid] < target) {
left = mid + 1;
} else {
right = mid - 1;
}
}
return -1;
}
const arr = [5, 7, 7, 8, 8, 10];
const target = 8;
console.log(searchRange(arr, target));
// [3, 4]
Complexity Analysis :
Time complexity: O(logn) Space complexity: O(1)
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