notes/OJ notes/pages/Leetcode Binary-Search.md

Leetcode Binary-Search

2022-07-09 09:34

Algorithms:

#algorithm #binary_search

Data structures:

#DS #array

Difficulty:

#coding_problem #difficulty-easy

Additional tags:

#leetcode #CS_list_need_practicing

Revisions:

N/A

Related topics:

tag:#binary_search

• Binary Search Algorithm
• Leetcode First-Bad-Version
• Leetcode Lowest-Common-Ancestor-Of-a-Binary-Search-Tree
• Leetcode Search-a-2D-Matrix
• Leetcode Search-Insert-Position
• Leetcode Two-Sum-IV-Input-Is-a-BST

Links:

• Link to problem

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Problem

Given an array of integers nums which is sorted in ascending order, and an integer target, write a function to search target in nums. If target exists, then return its index. Otherwise, return -1.

You must write an algorithm with O(log n) runtime complexity.

Examples

Example 1:

Input: nums = [-1,0,3,5,9,12], target = 9 Output: 4 Explanation: 9 exists in nums and its index is 4

Example 2:

Input: nums = [-1,0,3,5,9,12], target = 2 Output: -1 Explanation: 2 does not exist in nums so return -1

Constraints

• 1 <= nums.length <= 104
• -104 < nums[i], target < 104
• All the integers in nums are unique.
• nums is sorted in ascending order.

Thoughts

[!summary] This is a #binary_search problem.

Key takeout:

int r = nums.size() - 1;

make sure r is never OOB (l == r && r = array.size())

Solution

==#TODO: write in recursion==

iteration

class Solution {
public:
  int search(vector<int> &nums, int target) {
    // Why - 1?
    // Make sure mid is never OOB (l == r && r = size)
    int r = nums.size() - 1;
    int l = 0;
    int mid;
    int val;

    do {
      // l + ( r - l ) / 2 or (l + r) / 2 Both ok
      mid = (l + r) / 2;
      val = nums[mid];
      if (val == target) {
        return mid;
      } else if (val < target) {
        l = mid + 1;
      } else {
        r = mid - 1;
      }
    } while (l <= r);

    return -1;
  }
};