69 lines
1.7 KiB
C++
69 lines
1.7 KiB
C++
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/*
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Description: Given a sorted array of distinct integers and a target value,
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return the index if the target is found. If not, return the index where it
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would be if it were inserted in order.
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Approach: Using Binary search to solve in least time complexity.
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Time Complexity: O(log n)
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*/
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#include <iostream>
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#include <vector>
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using namespace std;
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//function
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int searchInsert(vector<int>& nums, int target) {
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//starting index
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int start=0;
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//ending index
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int end=nums.size()-1;
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while(start<=end){
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//calculating mid element
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int mid = start + (end-start)/2;
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//if target found, return the index
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if(nums[mid]==target){
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return mid;
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}
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if(nums[mid]>target){
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end=mid-1;
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}
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else{
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start=mid+1;
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}
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}
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//returning end+1 so that we can get the index where the target element can be inserted
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return end+1;
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}
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//main starts
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int main() {
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cout << "Enter number of elements in the array\n";
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int n;
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cin>>n;
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cout<<"Enter the array elements\n";
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vector<int> nums(n);
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for(int i=0;i<n;i++){
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cin>>nums[i];
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}
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cout<<"Enter the target element\n";
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int target;
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cin>>target;
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cout<<"The target element can be inserted at: "<<searchInsert(nums,target);
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return 0;
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}
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/*
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Example 1
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Input: nums = [1,3,5,6], target = 5
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Output: 2
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(Since the target element is present it will return its index)
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Example 2:
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Input: nums = [1,3,5,6], target = 2
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Output: 1
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(Since the target element is not present in the nums array, it will return the position where the target element can be inserted)
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We can insert 2 at position 1 - [1,2,3,5,6]
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*/
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