在排序数组中查找元素的第一个和最后一个 Position

Question

I am trying to solve the LeetCode problem Find First and Last Position of Element in Sorted Array :

Given an array of integers nums sorted in ascending 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.

I am writing this code:

var searchRange = function(nums, target) {
  nums = nums.sort((a, b) => (a-b))
  let result = [];
  for(let i = 0; i < nums.length; i++) {
    if (nums[i] == target) {
      result.push(i)
    } else {
     result = [-1, -1]
    }
  }

  return result
};
 
console.log(searchRange([5,7,7,8,8,10], 8));

It should return:

[3, 4]

but it returns:

[-1, -1]

What is wrong?

Answer 1

You loop does not have a good exit condition. Loop iterates till the last element and if the last element does not match, it sets -1,-1 into the result. Intermediate values of result are ignored.

Other problems:

• You're given a sorted array, so don't sort it again. That's heavy.
• When you iterate over all elements, it has O(n) time complexity.

Solution:

• Binary search to find any one occurence of the element - save the index where the value is found.
• Use the index found to check left and right adjacent elements in the original array to find the start and end positions.

Answer 2

Your else code is wiping out results from the previous iteration of the loop.

However, your code is not efficient:

First, it calls sort on an array that is given to be already sorted: so leave that out.

Secondly, as your code is visiting every element in the array, you still get a time complexity of O(n), not O(logn).

To get O(logn) you should implement a binary search, and then perform a search to find the start of the range, and another to find the end of it.

Here is how that could work:

 function binarySearch(nums, target) { let low = 0; let high = nums.length; while (low < high) { let mid = (low + high) >> 1; // half way if (nums[mid] > target) { high = mid; } else { low = mid + 1; } } return high; // first index AFTER target } function searchRange(nums, target) { let end = binarySearch(nums, target); let start = binarySearch(nums, target - 1); return start === end ? [-1, -1] : [start, end - 1]; } console.log(searchRange([5,7,7,8,8,10], 8));

Answer 3

 function searchRange(r,n){ var s = r.sort((a,b)=>ab); return [s.indexOf(n), s.lastIndexOf(n)]; } console.log('result 1',searchRange([5,7,7,8,8,10], 8)) console.log('result 2',searchRange([5,7,7,6,6,10], 8))

Answer 4

You were on the right track, its just your logic. You are setting result to [-1, -1] if ANY index of nums after the last target index is not equal to the target . So you want to have the check outside of the for loop.

var searchRange = function(nums, target) {
    nums = nums.sort((a, b) => (a-b));
    let result = [];
    for(let i = 0; i < nums.length; i++) {
        if (nums[i] == target) {
            result.push(i);
        }
    }

    return (result.length == 0) ? [-1, -1] : result;
};
 
 console.log(searchRange([5,7,7,8,8,10], 8));

Note: This will not be the final solution because this will return EVERY index that contains target

Answer 5

Dart

   if (nums.isEmpty || !nums.contains(target)) {
      return [-1, -1];
    } else {
      return [nums.indexOf(target), nums.lastIndexOf(target)];
    }