README_EN.md

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/0000-0099/0035.Search%20Insert%20Position/README_EN.md	Array Binary Search

35. Search Insert Position

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Description

Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

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

 

Example 1:

Input: nums = [1,3,5,6], target = 5
Output: 2

Example 2:

Input: nums = [1,3,5,6], target = 2
Output: 1

Example 3:

Input: nums = [1,3,5,6], target = 7
Output: 4

 

Constraints:

• 1 <= nums.length <= 104
• -104 <= nums[i] <= 104
• nums contains distinct values sorted in ascending order.
• -104 <= target <= 104

Solutions

Solution 1: Binary Search

Since the array $nums$ is already sorted, we can use the binary search method to find the insertion position of the target value $target$.

The time complexity is $O(\log n)$, and the space complexity is $O(1)$. Here, $n$ is the length of the array $nums$.

Python3

class Solution:
    def searchInsert(self, nums: List[int], target: int) -> int:
        l, r = 0, len(nums)
        while l < r:
            mid = (l + r) >> 1
            if nums[mid] >= target:
                r = mid
            else:
                l = mid + 1
        return l

Java

class Solution {
    public int searchInsert(int[] nums, int target) {
        int l = 0, r = nums.length;
        while (l < r) {
            int mid = (l + r) >>> 1;
            if (nums[mid] >= target) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}

C++

class Solution {
public:
    int searchInsert(vector<int>& nums, int target) {
        int l = 0, r = nums.size();
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums[mid] >= target) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
};

Go

func searchInsert(nums []int, target int) int {
	l, r := 0, len(nums)
	for l < r {
		mid := (l + r) >> 1
		if nums[mid] >= target {
			r = mid
		} else {
			l = mid + 1
		}
	}
	return l
}

TypeScript

function searchInsert(nums: number[], target: number): number {
    let [l, r] = [0, nums.length];
    while (l < r) {
        const mid = (l + r) >> 1;
        if (nums[mid] >= target) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return l;
}

Rust

impl Solution {
    pub fn search_insert(nums: Vec<i32>, target: i32) -> i32 {
        let mut l: usize = 0;
        let mut r: usize = nums.len();
        while l < r {
            let mid = (l + r) >> 1;
            if nums[mid] >= target {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        l as i32
    }
}

JavaScript

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var searchInsert = function (nums, target) {
    let [l, r] = [0, nums.length];
    while (l < r) {
        const mid = (l + r) >> 1;
        if (nums[mid] >= target) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return l;
};

PHP

class Solution {
    /**
     * @param Integer[] $nums
     * @param Integer $target
     * @return Integer
     */
    function searchInsert($nums, $target) {
        $l = 0;
        $r = count($nums);
        while ($l < $r) {
            $mid = $l + $r >> 1;
            if ($nums[$mid] >= $target) {
                $r = $mid;
            } else {
                $l = $mid + 1;
            }
        }
        return $l;
    }
}

Solution 2: Binary Search (Built-in Function)

We can also directly use the built-in function for binary search.

The time complexity is $O(\log n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.

Python3

class Solution:
    def searchInsert(self, nums: List[int], target: int) -> int:
        return bisect_left(nums, target)

Java

class Solution {
    public int searchInsert(int[] nums, int target) {
        int i = Arrays.binarySearch(nums, target);
        return i < 0 ? -i - 1 : i;
    }
}

C++

class Solution {
public:
    int searchInsert(vector<int>& nums, int target) {
        return lower_bound(nums.begin(), nums.end(), target) - nums.begin();
    }
};

Go

func searchInsert(nums []int, target int) int {
	return sort.SearchInts(nums, target)
}