//Counting Elements
//Given an integer array arr ,count element x such that x + 1 is also in arr
//lf there're duplicates in arr , count them seperately.
//Example 1:
//Input: arr =[1,2,3]
//0utput:2
//Explanation:1 and 2 are counted cause 2 and 3 are in arr.
//Example 2:
//Input: arr =[1,1,3,3,5,5,7,7]
//0utput:0
//Explanation: No numbers are counted, cause there's no 2,4,6, or 8 in arr.
//Example 3:
//Input: arr = [1,3,2,3,5,8]
//Output:3
//Explanation:0,1 and 2 are counted cause 1,2 and 3 are in arr.
//Example 4:
//Input: arr =[1,1,2,2]
//0utput:2
//Explanation:Two is are counted cause 2 is in arr.
cpp
#include <stdio.h>
int countElements(int *arr, int arrSize) {
// Boolean array to mark presence of elements in the input array.
int elementExists[1002] = {0}; // Supports values from 0 to 1000
// Mark the presence of elements in the array
for (int i = 0; i < arrSize; ++i) {
elementExists[arr[i]] = 1;
}
int returnNum = 0;
// Check if `arr[i] + 1` exists in the array
for (int i = 0; i < arrSize; ++i) {
if (elementExists[arr[i] + 1]) {
returnNum++;
}
}
return returnNum;
}
int main()
{
int arr[] = {1,1,2,2};
printf("%d",countElements(arr,sizeof(arr)/sizeof(arr[0])));
return 0;
}