Search insert position leetcode solution

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. 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 Example 4: Input: nums = [1,3,5,6], target = 0 Output: 0 Example 5: Input: nums = [1], target = 0 Output: 0 Constraints: • 1 = target) Python Solution class Solution: def searchInsert(self, nums: List[int], target: int) -> int: start = 0 end = len(nums) - 1 while start + 1 = target: return start if nums[end] >= target: return end return end + 1 • Time Complexity: O(log(N)) • Space Complexity: O(1) Posted in Post navigation

Table of Contents • • • • • • • • • • • • • • • • • • Example 1 2 3 4 5 Target = 3 2 1 3 5 7 9 Target = 8 4 Explanation Approach(Linear Search) We can run a loop to find the first index whose value exceeds or equals the target value. Proof: • If the value equals, then this index is the required index. • Target would have been inserted at this position, otherwise. Algorithm • Run a loop from first index to the end of the array • If array[i] exceeds or equals target value, then return i • return n, as we found no index such that array[index] >= target, so target should be inserted in the end • Print the result Implementation of algorithm to find Search Insert Position of Target C++ Program #include using namespace std; int searchInsert(vector& a, int target) 4 Complexity Analysis of finding Search Insert Position of Target Leetcode Solution Time Complexity O(logN). In the worst case, we can make logN comparisons. Space complexity O(1). Again, we use constant space for variables. Categories Tags Post navigation

• 1 Intuition: • We have given a sorted array of distinct integers and a target value. • We have to find the position where target value should inserted. • The very first intuition is this only that we can travel in the array and then find the position. Brute Force: • Since, question ask us to find the solution in O(log n) time complexity but we can hit a try of solving it in some basic method. • We know that array is sorted, so we can travel in the array and then till ith element of the array is less than target, we will move forward. • After a point we found out that ith element of array is either greater or equals to the target value, we stop. • And that should be our answer. Time And Space Complexity: • Time Complexity: O(n), since we are travelling in the array. • Space Complexity: O(1), we are not using data structure from our side. class Solution ; Hopefully, you guys understand the solution, Thanks to for contributing this solution.

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. 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 Example 4: Input: nums = [1,3,5,6], target = 0 Output: 0 Example 5: Input: nums = [1], target = 0 Output: 0 Constraints: • 1 = target) Python Solution class Solution: def searchInsert(self, nums: List[int], target: int) -> int: start = 0 end = len(nums) - 1 while start + 1 = target: return start if nums[end] >= target: return end return end + 1 • Time Complexity: O(log(N)) • Space Complexity: O(1) Posted in Post navigation

• 1 Intuition: • We have given a sorted array of distinct integers and a target value. • We have to find the position where target value should inserted. • The very first intuition is this only that we can travel in the array and then find the position. Brute Force: • Since, question ask us to find the solution in O(log n) time complexity but we can hit a try of solving it in some basic method. • We know that array is sorted, so we can travel in the array and then till ith element of the array is less than target, we will move forward. • After a point we found out that ith element of array is either greater or equals to the target value, we stop. • And that should be our answer. Time And Space Complexity: • Time Complexity: O(n), since we are travelling in the array. • Space Complexity: O(1), we are not using data structure from our side. class Solution ; Hopefully, you guys understand the solution, Thanks to for contributing this solution.

Table of Contents • • • • • • • • • • • • • • • • • • Example 1 2 3 4 5 Target = 3 2 1 3 5 7 9 Target = 8 4 Explanation Approach(Linear Search) We can run a loop to find the first index whose value exceeds or equals the target value. Proof: • If the value equals, then this index is the required index. • Target would have been inserted at this position, otherwise. Algorithm • Run a loop from first index to the end of the array • If array[i] exceeds or equals target value, then return i • return n, as we found no index such that array[index] >= target, so target should be inserted in the end • Print the result Implementation of algorithm to find Search Insert Position of Target C++ Program #include using namespace std; int searchInsert(vector& a, int target) 4 Complexity Analysis of finding Search Insert Position of Target Leetcode Solution Time Complexity O(logN). In the worst case, we can make logN comparisons. Space complexity O(1). Again, we use constant space for variables. Categories Tags Post navigation

• 1 Intuition: • We have given a sorted array of distinct integers and a target value. • We have to find the position where target value should inserted. • The very first intuition is this only that we can travel in the array and then find the position. Brute Force: • Since, question ask us to find the solution in O(log n) time complexity but we can hit a try of solving it in some basic method. • We know that array is sorted, so we can travel in the array and then till ith element of the array is less than target, we will move forward. • After a point we found out that ith element of array is either greater or equals to the target value, we stop. • And that should be our answer. Time And Space Complexity: • Time Complexity: O(n), since we are travelling in the array. • Space Complexity: O(1), we are not using data structure from our side. class Solution ; Hopefully, you guys understand the solution, Thanks to for contributing this solution.

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. 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 Example 4: Input: nums = [1,3,5,6], target = 0 Output: 0 Example 5: Input: nums = [1], target = 0 Output: 0 Constraints: • 1 = target) Python Solution class Solution: def searchInsert(self, nums: List[int], target: int) -> int: start = 0 end = len(nums) - 1 while start + 1 = target: return start if nums[end] >= target: return end return end + 1 • Time Complexity: O(log(N)) • Space Complexity: O(1) Posted in Post navigation

Table of Contents • • • • • • • • • • • • • • • • • • Example 1 2 3 4 5 Target = 3 2 1 3 5 7 9 Target = 8 4 Explanation Approach(Linear Search) We can run a loop to find the first index whose value exceeds or equals the target value. Proof: • If the value equals, then this index is the required index. • Target would have been inserted at this position, otherwise. Algorithm • Run a loop from first index to the end of the array • If array[i] exceeds or equals target value, then return i • return n, as we found no index such that array[index] >= target, so target should be inserted in the end • Print the result Implementation of algorithm to find Search Insert Position of Target C++ Program #include using namespace std; int searchInsert(vector& a, int target) 4 Complexity Analysis of finding Search Insert Position of Target Leetcode Solution Time Complexity O(logN). In the worst case, we can make logN comparisons. Space complexity O(1). Again, we use constant space for variables. Categories Tags Post navigation

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. 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 Example 4: Input: nums = [1,3,5,6], target = 0 Output: 0 Example 5: Input: nums = [1], target = 0 Output: 0 Constraints: • 1 = target) Python Solution class Solution: def searchInsert(self, nums: List[int], target: int) -> int: start = 0 end = len(nums) - 1 while start + 1 = target: return start if nums[end] >= target: return end return end + 1 • Time Complexity: O(log(N)) • Space Complexity: O(1) Posted in Post navigation