CODEWITHSUNDEEP
carraysfunctionfor-loop
muhammed kartal
muhammed kartal

Reputation: 47

Smallest odd number in given array

This code is supposed to find the smallest odd number in given array and store it in min but when I try to print min it always prints 0.

int smallestodd(int x[5]){
  int j;
  int k[5];
  int p = 0;
  int r = 0;

  for(int h =0; h<5;h++){

    j = x[h] % 2;
    if(j == 1){
      int temp =x[h];
      k[p] =temp;
      p++;
    }
  }

  int min = k[0];


  while(k[r] !=0){
    if(k[r] < min ){
      min = k[r];
      r++;
    }
  }

  return min;
}

Upvotes: 4

Views: 6809

Answers (4)

COBIEN
COBIEN

Reputation: 1

Use this using statement as first :

Using System Linq;

Console.WriteLine(myArray.Where(i => i%2 == 1).Min());

Upvotes: 0

pmg
pmg

Reputation: 108988

Assuming there is an odd number in the array -- let's say trying to find the minimum odd number in an array with just even numbers (or no numbers) is UB :)

index = 0;
while (arr[index] % 2 == 0) index++; // skip even numbers
min = arr[index++]; // first odd number
while (index < length) {
    if (arr[index] % 2) {
        if (arr[index] < min) min = arr[index];
    }
    index++;
}

Upvotes: 5

seduardo
seduardo

Reputation: 704

this code avoid overflow in search and return 1 when found or 0 if array has only even numbers.

int getMinOdd(int arr[], int length, int *value) {
    int found = 0;
    for(int idx=0; idx < length; idx++) {
        if (arr[idx] % 2) {
            if (!found || *value > arr[idx]) {
                *value = arr[idx];
            }
            found = 1;
        }
    }
    return found;
}

Upvotes: 1

Karan
Karan

Reputation: 74

It's quite simple actually. You need to just check 2 conditions on your array.

int smallestOdd(int arr[]){
    int min = 99999; //Some very large number
    for(int i = 0; i < (length of your array); i++) {
        if(arr[i]%2 != 0 && arr[i] < min) { //Check if number is odd and less than the current minimum value
            min = arr[i];
        }
    }
    return min;
}

Upvotes: 0

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