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++
#include <bits/stdc++.h>
using namespace std;
int find_index(int arr[], int n, int K)
{
for (int i = 0; i < n; i++)
if (arr[i] == K)
return i;
else if (arr[i] > K)
return i;
return n;
}
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;
}
|
C
#include <stdio.h>
int find_index(int arr[], int n, int K)
{
for (int i = 0; i < n; i++)
if (arr[i] == K)
return i;
else if (arr[i] > K)
return i;
return n;
}
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;
}
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Java
import java.io.*;
class GFG{
static int find_index(int[] arr, int n, int K)
{
for(int i = 0; i < n; i++)
if (arr[i] == K)
return i;
else if (arr[i] > K)
return i;
return n;
}
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));
}
}
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Python3
def find_index(arr, n, K):
for i in range(n):
if arr[i] == K:
return i
elif arr[i] > K:
return i
return n
arr = [1, 3, 5, 6]
n = len(arr)
K = 2
print(find_index(arr, n, K))
|
C#
using System;
class GFG{
static int find_index(int[] arr, int n, int K)
{
for(int i = 0; i < n; i++)
if (arr[i] == K)
return i;
else if (arr[i] > K)
return i;
return n;
}
public static void Main()
{
int[] arr = { 1, 3, 5, 6 };
int n = arr.Length;
int K = 2;
Console.WriteLine(find_index(arr, n, K));
}
}
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Javascript
<script>
function find_index(arr, n, K)
{
for(let i = 0; i < n; i++)
if (arr[i] == K)
return i;
else if (arr[i] > K)
return i;
return n;
}
let arr = [ 1, 3, 5, 6 ];
let n = arr.length;
let K = 2;
document.write(find_index(arr, n, K));
</script>
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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++
#include <bits/stdc++.h>
using namespace std;
int find_index(int arr[], int n, int K)
{
int start = 0;
int end = n - 1;
while (start <= end) {
int mid = (start + end) / 2;
if (arr[mid] == K)
return mid;
else if (arr[mid] < K)
start = mid + 1;
else
end = mid - 1;
}
return end + 1;
}
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;
}
|
C
#include<stdio.h>
int find_index(int arr[], int n, int K)
{
int start = 0;
int end = n - 1;
while (start <= end) {
int mid = (start + end) / 2;
if (arr[mid] == K)
return mid;
else if (arr[mid] < K)
start = mid + 1;
else
end = mid - 1;
}
return end + 1;
}
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;
}
|
Java
import java.io.*;
class GFG{
static int find_index(int[] arr, int n, int K)
{
int start = 0;
int end = n - 1;
while (start <= end)
{
int mid = (start + end) / 2;
if (arr[mid] == K)
return mid;
else if (arr[mid] < K)
start = mid + 1;
else
end = mid - 1;
}
return end + 1;
}
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));
}
}
|
Python3
def find_index(arr, n, K):
start = 0
end = n - 1
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 end + 1
arr = [1, 3, 5, 6]
n = len(arr)
K = 2
print(find_index(arr, n, K))
|
C#
using System;
class GFG{
static int find_index(int[] arr, int n, int K)
{
int start = 0;
int end = n - 1;
while (start <= end)
{
int mid = (start + end) / 2;
if (arr[mid] == K)
return mid;
else if (arr[mid] < K)
start = mid + 1;
else
end = mid - 1;
}
return end + 1;
}
public static void Main()
{
int[] arr = { 1, 3, 5, 6 };
int n = arr.Length;
int K = 2;
Console.WriteLine(find_index(arr, n, K));
}
}
|
Javascript
<script>
function find_index(arr, n, K)
{
let start = 0;
let end = n - 1;
while (start <= end) {
let mid = Math.floor((start + end) / 2);
if (arr[mid] == K)
return mid;
else if (arr[mid] < K)
start = mid + 1;
else
end = mid - 1;
}
return end + 1;
}
let arr = [ 1, 3, 5, 6 ];
let n = arr.length;
let K = 2;
document.write(find_index(arr, n, K) + "<br>");
</script>
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Time Complexity: O(log N)
Auxiliary Space: O(1)