Count of elements A[i] such that A[i] + 1 is also present in the Array
Last Updated :
12 Jul, 2025
Given an integer array arr the task is to count the number of elements 'A[i]', such that A[i] + 1 is also present in the array.
Note: If there are duplicates in the array, count them separately.
Examples:
Input: arr = [1, 2, 3]
Output: 2
Explanation:
1 and 2 are counted cause 2 and 3 are in arr.
Input: arr = [1, 1, 3, 3, 5, 5, 7, 7]
Output: 0
Approach 1: Brute Force Solution
For all the elements in the array, return the total count after examining all elements
- For current element x, compute x + 1, and search all positions before and after the current value for x + 1.
- If you find x + 1, add 1 to the total count
Below is the implementation of the above approach:
C++
// C++ program to count of elements
// A[i] such that A[i] + 1
// is also present in the Array
#include <bits/stdc++.h>
using namespace std;
// Function to find the countElements
int countElements(int* arr, int n)
{
// Initialize count as zero
int count = 0;
// Iterate over each element
for (int i = 0; i < n; i++) {
// Store element in int x
int x = arr[i];
// Calculate x + 1
int xPlusOne = x + 1;
// Initialize found as false
bool found = false;
// Run loop to search for x + 1
// after the current element
for (int j = i + 1; j < n; j++) {
if (arr[j] == xPlusOne) {
found = true;
break;
}
}
// Run loop to search for x + 1
// before the current element
for (int k = i - 1;
!found && k >= 0; k--) {
if (arr[k] == xPlusOne) {
found = true;
break;
}
}
// if found is true, increment count
if (found == true) {
count++;
}
}
return count;
}
// Driver program
int main()
{
int arr[] = { 1, 2, 3 };
int n = sizeof(arr) / sizeof(arr[0]);
// call countElements function on array
cout << countElements(arr, n);
return 0;
}
// Java program to count of elements
// A[i] such that A[i] + 1
// is also present in the Array
import java.io.*;
import java.util.Arrays;
class GFG{
// Function to find the countElements
public static int countElements(int[] arr, int n)
{
// Initialize count as zero
int count = 0;
// Iterate over each element
for (int i = 0; i < n; i++)
{
// Store element in int x
int x = arr[i];
// Calculate x + 1
int xPlusOne = x + 1;
// Initialize found as false
boolean found = false;
// Run loop to search for x + 1
// after the current element
for (int j = i + 1; j < n; j++)
{
if (arr[j] == xPlusOne)
{
found = true;
break;
}
}
// Run loop to search for x + 1
// before the current element
for (int k = i - 1; !found && k >= 0; k--)
{
if (arr[k] == xPlusOne)
{
found = true;
break;
}
}
// If found is true, increment count
if (found == true)
{
count++;
}
}
return count;
}
// Driver code
public static void main (String[] args)
{
int arr[] = { 1, 2, 3 };
int n = arr.length;
// Call countElements function on array
System.out.println(countElements(arr, n));
}
}
//This code is contributed by shubhamsingh10
# Python3 program to count of elements
# A[i] such that A[i] + 1
# is also present in the Array
# Function to find the countElements
def countElements(arr,n):
# Initialize count as zero
count = 0
# Iterate over each element
for i in range(n):
# Store element in int x
x = arr[i]
# Calculate x + 1
xPlusOne = x + 1
# Initialize found as false
found = False
# Run loop to search for x + 1
# after the current element
for j in range(i + 1,n,1):
if (arr[j] == xPlusOne):
found = True
break
# Run loop to search for x + 1
# before the current element
k = i - 1
while(found == False and k >= 0):
if (arr[k] == xPlusOne):
found = True
break
k -= 1
# if found is true, increment count
if (found == True):
count += 1
return count
# Driver program
if __name__ == '__main__':
arr = [1, 2, 3]
n = len(arr)
# call countElements function on array
print(countElements(arr, n))
# This code is contributed by Surendra_Gangwar
// C# program to count of elements
// A[i] such that A[i] + 1
// is also present in the Array
using System;
class GFG{
// Function to find the countElements
static int countElements(int[] arr, int n)
{
// Initialize count as zero
int count = 0;
// Iterate over each element
for (int i = 0; i < n; i++) {
// Store element in int x
int x = arr[i];
// Calculate x + 1
int xPlusOne = x + 1;
// Initialize found as false
bool found = false;
// Run loop to search for x + 1
// after the current element
for (int j = i + 1; j < n; j++) {
if (arr[j] == xPlusOne) {
found = true;
break;
}
}
// Run loop to search for x + 1
// before the current element
for (int k = i - 1;
!found && k >= 0; k--) {
if (arr[k] == xPlusOne) {
found = true;
break;
}
}
// if found is true,
// increment count
if (found == true) {
count++;
}
}
return count;
}
// Driver program
static public void Main ()
{
int[] arr = { 1, 2, 3 };
int n = arr.Length;
// call countElements function on array
Console.WriteLine(countElements(arr, n));
}
}
// This code is contributed by shubhamsingh10
<script>
// JavaScript program to count of elements
// A[i] such that A[i] + 1
// is also present in the Array
// Function to find the countElements
function countElements(arr, n)
{
// Initialize count as zero
let count = 0;
// Iterate over each element
for (let i = 0; i < n; i++) {
// Store element in int x
let x = arr[i];
// Calculate x + 1
let xPlusOne = x + 1;
// Initialize found as false
let found = false;
// Run loop to search for x + 1
// after the current element
for (let j = i + 1; j < n; j++)
{
if (arr[j] == xPlusOne)
{
found = true;
break;
}
}
// Run loop to search for x + 1
// before the current element
for (let k = i - 1;
!found && k >= 0; k--) {
if (arr[k] == xPlusOne) {
found = true;
break;
}
}
// if found is true, increment count
if (found == true) {
count++;
}
}
return count;
}
// Driver program
let arr = [ 1, 2, 3 ];
let n = arr.length;
// call countElements function on array
document.write(countElements(arr, n));
// This code is contributed by Surbhi Tyagi.
</script>
Time Complexity: In the above approach, for a given element, we check all other elements, So the time complexity is O(N*N) where N is no of elements.
Auxiliary Space Complexity: In the above approach, we are not using any additional space, so Auxiliary space complexity is O(1).
Approach 2: Using Map
- For all elements in the array, say x, add x-1 to the map
- Again, for all elements in the array, say x, check if it exists in the map. If it exists, increment the counter
- Return the total count after examining all keys in the map
Below is the implementation of the above approach:
C++
// C++ program to count of elements
// A[i] such that A[i] + 1
// is also present in the Array
#include <bits/stdc++.h>
using namespace std;
// Function to find the countElements
int countElements(vector<int>& arr)
{
int size = arr.size();
// Initialize result as zero
int res = 0;
// Create map
map<int, bool> dat;
// First loop to fill the map
for (int i = 0; i < size; ++i) {
dat[arr[i] - 1] = true;
}
// Second loop to check the map
for (int i = 0; i < size; ++i) {
if (dat[arr[i]] == true) {
res++;
}
}
return res;
}
// Driver program
int main()
{
// Input Array
vector<int> arr = { 1, 3, 2, 3, 5, 0 };
// Call the countElements function
cout << countElements(arr) << endl;
return 0;
}
// Java program to count of elements
// A[i] such that A[i] + 1 is
// also present in the Array
import java.util.*;
class GFG{
// Function to find the countElements
public static int countElements(int[] arr)
{
int size = arr.length;
// Initialize result as zero
int res = 0;
// Create map
Map<Integer, Boolean> dat = new HashMap<>();
// First loop to fill the map
for(int i = 0; i < size; ++i)
{
dat.put((arr[i] - 1), true);
}
// Second loop to check the map
for(int i = 0; i < size; ++i)
{
if (dat.containsKey(arr[i]) == true)
{
res++;
}
}
return res;
}
// Driver code
public static void main(String[] args)
{
// Input Array
int[] arr = { 1, 3, 2, 3, 5, 0 };
// Call the countElements function
System.out.println(countElements(arr));
}
}
// This code is contributed by shad0w1947
# Python program to count of elements
# A[i] such that A[i] + 1
# is also present in the Array
# Function to find the countElements
def countElements(arr):
size = len(arr)
# Initialize result as zero
res = 0
# Create map
dat={}
# First loop to fill the map
for i in range(size):
dat[arr[i] - 1] = True
# Second loop to check the map
for i in range(size):
if (arr[i] in dat):
res += 1
return res
# Driver program
# Input Array
arr = [1, 3, 2, 3, 5, 0]
# Call the countElements function
print(countElements(arr))
# This code is contributed by shubhamsingh10
// C# program to count of elements
// A[i] such that A[i] + 1 is
// also present in the Array
using System;
using System.Collections.Generic;
class GFG{
// Function to find the countElements
public static int countElements(int[] arr)
{
int size = arr.Length;
// Initialize result as zero
int res = 0;
// Create map
Dictionary<int,
Boolean> dat = new Dictionary<int,
Boolean>();
// First loop to fill the map
for(int i = 0; i < size; ++i)
{
if(!dat.ContainsKey(arr[i] - 1))
dat.Add((arr[i] - 1), true);
}
// Second loop to check the map
for(int i = 0; i < size; ++i)
{
if (dat.ContainsKey(arr[i]) == true)
{
res++;
}
}
return res;
}
// Driver code
public static void Main(String[] args)
{
// Input Array
int[] arr = { 1, 3, 2, 3, 5, 0 };
// Call the countElements function
Console.WriteLine(countElements(arr));
}
}
// This code is contributed by 29AjayKumar
<script>
// javascript program to count of elements
// A[i] such that A[i] + 1 is
// also present in the Array
// Function to find the countElements
function countElements(arr) {
var size = arr.length;
// Initialize result as zero
var res = 0;
// Create map
var dat = new Map();
// First loop to fill the map
for (i = 0; i < size; ++i) {
dat.set((arr[i] - 1), true);
}
// Second loop to check the map
for (i = 0; i < size; ++i) {
if (dat.has(arr[i]) == true) {
res++;
}
}
return res;
}
// Input Array
var arr = [ 1, 3, 2, 3, 5, 0 ];
// Call the countElements function
document.write(countElements(arr));
// This code is contributed by umadevi9616
</script>
Time Complexity: In the above approach, we iterate over the array twice. Once for filling the map and second time for checking the elements in the map, So the time complexity is O(N) where N is no of elements.
Auxiliary Space Complexity: In the above approach, we are using an additional map which can contain N elements, so auxiliary space complexity is O(N).
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