LeetCode #34: Find First and Last Position of Element in Sorted Array (C/C++).

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source: https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/
C/C++ Solution to LeetCode problem 34. Find First and Last Position of Element in Sorted Array.

Problem

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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.

Examples

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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

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• 0 <= nums.length <= 105
• -109 <= nums[i] <= 109
• nums is a non-decreasing array.
• -109 <= target <= 109

Solution

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We perform two binary searches.

• Binary search to find the minimum index of the target value.
• If there is a minimun index, then we perform a binary search to find the maximum index.
  • Because there is a minimum, we can initialize the left pointer to the first position where the target was found for the first time.

class Solution {
public:
  vector<int> searchRange(vector<int>& nums, int target) {
    int start = -1;
    int end = -1;

    int l = 0;
    int r = nums.size() - 1;
    while (l<=r) {
      int m = (l + r) / 2;
      if (nums[m] == target) {
        start = m;
        end = max(end, m);
      }
      if (nums[m] >= target)
        r = m - 1;
      else
        l = m + 1;
    }

    if (end != -1) {
      l = end;
      r = nums.size() - 1;
      while (l<=r) {
        int m = (l + r) / 2;
        if (nums[m] == target)
          end = max(end, m);
        if (nums[m] <= target)
          l = m + 1;
        else
          r = m - 1;
      }
    }

    return {start, end};
  }
};