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# Search insert position of K in a sorted array

• Last Updated : 07 May, 2021

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 = 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 ``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);``    ``int` `K = 2;``    ``cout << find_index(arr, n, K) << endl;``    ``return` `0;``}`

## 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

 ``
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 low = mid + 1. Otherwise, set high = mid – 1.

Below is the implementation of above approach :

## C++

 `// C++ program for the above approach` `#include ``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);``    ``int` `K = 2;``    ``cout << find_index(arr, n, K) << endl;``    ``return` `0;``}`

## 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

 ``
Output:
`1`

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

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