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Search insert position of K in a sorted array

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  • Difficulty Level : Medium
  • Last Updated : 24 Jun, 2022
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Given a sorted array arr[] consisting of N distinct integers and an integer K, the task is to find the index of K, if it’s present in the array arr[]. Otherwise, find the index where K must be inserted to keep the array sorted.

Examples: 

Input: arr[] = {1, 3, 5, 6}, K = 5
Output: 2
Explanation: Since 5 is found at index 2 as arr[2] = 5, the output is 2.

Input: arr[] = {1, 3, 5, 6}, K = 2
Output: 1
Explanation: Since 2 is not present in the array but can be inserted at index 1 to make the array sorted.

Naive Approach: Follow the steps below to solve the problem:

  • Iterate over every element of the array arr[] and search for integer K.
  • If any array element is found to be equal to K, then print index of K.
  • Otherwise, if any array element is found to be greater than K, print that index as the insert position of K. If no element is found to be exceeding K, K must be inserted after the last array element.

Below is the implementation of above approach :

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find insert position of K
int find_index(int arr[], int n, int K)
{
    // Traverse the array
    for (int i = 0; i < n; i++)
        // If K is found
        if (arr[i] == K)
            return i;
        // If current array element exceeds K
        else if (arr[i] > K)
            return i;
    // If all elements are smaller than K
    return n;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 3, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int K = 2;
    cout << find_index(arr, n, K) << endl;
    return 0;
}
 
// This code is contributed by Aditya Kumar (adityakumar129)

C




// C program for the above approach
#include <stdio.h>
 
// Function to find insert position of K
int find_index(int arr[], int n, int K)
{
    // Traverse the array
    for (int i = 0; i < n; i++)
        // If K is found
        if (arr[i] == K)
            return i;
        // If current array element exceeds K
        else if (arr[i] > K)
            return i;
    // If all elements are smaller than K
    return n;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 3, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int K = 2;
    printf("%d\n", find_index(arr, n, K));
    return 0;
}
 
// This code is contributed by Aditya Kumar (adityakumar129)

Java




// Java program for the above approach
import java.io.*;
 
class GFG{
 
// Function to find insert position of K
static int find_index(int[] arr, int n, int K)
{
     
    // Traverse the array
    for(int i = 0; i < n; i++)
     
        // If K is found
        if (arr[i] == K)
            return i;
 
        // If current array element
        // exceeds K
        else if (arr[i] > K)
            return i;
 
    // If all elements are smaller
    // than K
    return n;
}
 
// Driver Code
public static void main(String[] args)
{
    int[] arr = { 1, 3, 5, 6 };
    int n = arr.length;
    int K = 2;
     
    System.out.println(find_index(arr, n, K));
}
}
 
// This code is contributed by akhilsaini

Python3




# Python program for the above approach
 
# Function to find insert position of K
def find_index(arr, n, K):
     
    # Traverse the array
    for i in range(n):
         
        # If K is found
        if arr[i] == K:
            return i
             
        # If arr[i] exceeds K
        elif arr[i] > K:
            return i
             
    # If all array elements are smaller
    return n
 
# Driver Code
arr = [1, 3, 5, 6]
n = len(arr)
K = 2
print(find_index(arr, n, K))

C#




// C# program for the above approach
using System;
 
class GFG{
 
// Function to find insert position of K
static int find_index(int[] arr, int n, int K)
{
     
    // Traverse the array
    for(int i = 0; i < n; i++)
     
        // If K is found
        if (arr[i] == K)
            return i;
 
        // If current array element
        // exceeds K
        else if (arr[i] > K)
            return i;
 
    // If all elements are smaller
    // than K
    return n;
}
 
// Driver Code
public static void Main()
{
    int[] arr = { 1, 3, 5, 6 };
    int n = arr.Length;
    int K = 2;
     
    Console.WriteLine(find_index(arr, n, K));
}
}
 
// This code is contributed by akhilsaini

Javascript




<script>
 
// Javascript program for the above approach
 
// Function to find insert position of K
function find_index(arr, n, K)
{
     
    // Traverse the array
    for(let i = 0; i < n; i++)
      
        // If K is found
        if (arr[i] == K)
            return i;
  
        // If current array element
        // exceeds K
        else if (arr[i] > K)
            return i;
  
    // If all elements are smaller
    // than K
    return n;
}
 
// Driver code
let arr = [ 1, 3, 5, 6 ];
let n = arr.length;
let K = 2;
  
document.write(find_index(arr, n, K));
 
// This code is contributed by splevel62  
 
</script>

Output: 

1

 

Time Complexity: O(N)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to use Binary Search. Follow the steps below to solve the problem: 

  • Set start and end as 0 and N – 1, where the start and end variables denote the lower and upper bound of the search space respectively.
  • Calculate mid = (start + end) / 2.
  • If arr[mid] is found to be equal to K, print mid as the required answer.
  • If arr[mid] exceeds K, set high = mid – 1 Otherwise, set  low = mid + 1

Below is the implementation of above approach :

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find insert position of K
int find_index(int arr[], int n, int K)
{
    // Lower and upper bounds
    int start = 0;
    int end = n - 1;
    // Traverse the search space
    while (start <= end) {
        int mid = (start + end) / 2;
        // If K is found
        if (arr[mid] == K)
            return mid;
        else if (arr[mid] < K)
            start = mid + 1;
        else
            end = mid - 1;
    }
    // Return insert position
    return end + 1;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 3, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int K = 2;
    cout << find_index(arr, n, K) << endl;
    return 0;
}
 
// This code is contributed by Aditya Kumar (adityakumar129)

C




// C program for the above approach
 
#include<stdio.h>
 
// Function to find insert position of K
int find_index(int arr[], int n, int K)
{
    // Lower and upper bounds
    int start = 0;
    int end = n - 1;
    // Traverse the search space
    while (start <= end) {
        int mid = (start + end) / 2;
        // If K is found
        if (arr[mid] == K)
            return mid;
        else if (arr[mid] < K)
            start = mid + 1;
        else
            end = mid - 1;
    }
    // Return insert position
    return end + 1;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 3, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int K = 2;
    printf("%d",find_index(arr, n, K));
    return 0;
}
 
// This code is contributed by Aditya Kumar (adityakumar129)

Java




// Java program for the above approach
import java.io.*;
 
class GFG{
 
// Function to find insert position of K
static int find_index(int[] arr, int n, int K)
{
     
    // Lower and upper bounds
    int start = 0;
    int end = n - 1;
 
    // Traverse the search space
    while (start <= end)
    {
        int mid = (start + end) / 2;
 
        // If K is found
        if (arr[mid] == K)
            return mid;
 
        else if (arr[mid] < K)
            start = mid + 1;
 
        else
            end = mid - 1;
    }
 
    // Return insert position
    return end + 1;
}
 
// Driver Code
public static void main(String[] args)
{
    int[] arr = { 1, 3, 5, 6 };
    int n = arr.length;
    int K = 2;
     
    System.out.println(find_index(arr, n, K));
}
}
 
// This code is contributed by akhilsaini

Python3




# Python program to implement
# the above approach
 
# Function to find insert position of K
def find_index(arr, n, B):
 
    # Lower and upper bounds
    start = 0
    end = n - 1
 
    # Traverse the search space
    while start<= end:
 
        mid =(start + end)//2
 
        if arr[mid] == K:
            return mid
 
        elif arr[mid] < K:
            start = mid + 1
        else:
            end = mid-1
 
    # Return the insert position
    return end + 1
 
# Driver Code
arr = [1, 3, 5, 6]
n = len(arr)
K = 2
print(find_index(arr, n, K))

C#




// C# program for the above approach
using System;
 
class GFG{
 
// Function to find insert position of K
static int find_index(int[] arr, int n, int K)
{
     
    // Lower and upper bounds
    int start = 0;
    int end = n - 1;
 
    // Traverse the search space
    while (start <= end)
    {
        int mid = (start + end) / 2;
 
        // If K is found
        if (arr[mid] == K)
            return mid;
 
        else if (arr[mid] < K)
            start = mid + 1;
 
        else
            end = mid - 1;
    }
 
    // Return insert position
    return end + 1;
}
 
// Driver Code
public static void Main()
{
    int[] arr = { 1, 3, 5, 6 };
    int n = arr.Length;
    int K = 2;
     
    Console.WriteLine(find_index(arr, n, K));
}
}
 
// This code is contributed by akhilsaini

Javascript




<script>
// JavaScript program for the above approach
 
// Function to find insert position of K
function find_index(arr, n, K)
{
    // Lower and upper bounds
    let start = 0;
    let end = n - 1;
 
    // Traverse the search space
    while (start <= end) {
        let mid = Math.floor((start + end) / 2);
 
        // If K is found
        if (arr[mid] == K)
            return mid;
 
        else if (arr[mid] < K)
            start = mid + 1;
 
        else
            end = mid - 1;
    }
 
    // Return insert position
    return end + 1;
}
 
// Driver Code
    let arr = [ 1, 3, 5, 6 ];
    let n = arr.length;
    let K = 2;
    document.write(find_index(arr, n, K) + "<br>");
 
 
 
 
// This code is contributed by Surbhi Tyagi.
</script>

Output: 

1

 

Time Complexity: O(log N)
Auxiliary Space: O(1)


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