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C Exercises: Find the number occurring odd number of times in an array

C Array: Exercise-34 with Solution

Write a program in C to find the number occurring odd number of times in an array.
All numbers occur even number of times except one number which occurs odd number of times.

Pictorial Presentation:

C Exercises: Find the number occurring odd number of times in an array

Sample Solution:

C Code:

#include <stdio.h>
 
int findOddCountElem (int *arr1, int n ) 
{
     int i, ResultXor = 0; 
     for(i = 0; i < n; i++)
	 {
      ResultXor = ResultXor ^ arr1[i];
	  }
     return ResultXor;
}
 
int main()
{
    int i;
 int arr1[] = {8, 3, 8, 5, 4, 3, 4, 3, 5};
 
    int ctr = sizeof(arr1)/sizeof(arr1[0]);
    printf("The given array is :  ");

	for(i = 0; i < ctr; i++)
	{
	printf("%d  ", arr1[i]);
    } 
    printf("\n");
 
 printf("Number of odd number occur(s) : %d times.\n", findOddCountElem(arr1, ctr));
 return 0;
}

Sample Output: 

The given array is :  8  3  8  5  4  3  4  3  5  
Number of odd number occur(s) : 3 times.

Flowchart :

Flowchart: Find the number occurring odd number of times in an array.

C Programming Code Editor:

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Previous: Write a program in C to find the majority element of an array.
Next: Write a program in C to find the largest sum of contiguous subarray of an array.

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C Programming: Tips of the Day

Returning an array using C

You can't return arrays from functions in C. You also can't (shouldn't) do this:

char *returnArray(char array []){
 char returned [10];
 //methods to pull values from array, interpret them, and then create new array
 return &(returned[0]); //is this correct?
}

returned is created with automatic storage duration and references to it will become invalid once it leaves its declaring scope, i.e., when the function returns.

You will need to dynamically allocate the memory inside of the function or fill a preallocated buffer provided by the caller.

Dynamically allocate the memory inside of the function (caller responsible for deallocating ret)

char *foo(int count) {
    char *ret = malloc(count);
    if(!ret)
        return NULL;

    for(int i = 0; i < count; ++i) 
        ret[i] = i;

    return ret;
}

Call it like so:

int main() {
    char *p = foo(10);
    if(p) {
        // do stuff with p
        free(p);
    }

    return 0;
}

Ref : https://bit.ly/3yFIeao

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