The Highest Even Number in a Array

function longestEven(arr) {
  let firstOdd = -1,
    lastOdd = -1,
    oddCount = 0;
  for (let i = 0; i < arr.length; i++) {
    if (arr[i] % 2 === 1) {
      if (firstOdd === -1) firstOdd = i;
      lastOdd = i;
      ++oddCount;
    }
  }
  return oddCount % 2 === 0 ? arr.length : 
    Math.max(arr.length - (firstOdd + 1), lastOdd);
}
console.log(longestEven([2, 3, 1, 8, 4, 7, 2]));

Find the Min Even Integer and Max Even Integer in an Array

This test case exposes one of the dark sides of your code where you have initialized your min and max to A[1] regardless of whether A[1] is odd or even.

Problem:

There may be cases where the array does not contain any even number. In such cases, you may wish to print so instead of printing -1 or INT_MIN or INT_MAX.

If you are new to INT_MIN and INT_MAX, this will serve as a starter before proceeding to the solution.

Solution:

This solution modifies your code in such a way that it handles all the cases given that you provide the inputs without any error.

1. Have a flag to know whether you found your min and max:

   int foundAnswer = 0;
   

2. Add the header file limits.h and initialize your min and max as
   follows:

   min = INT_MAX;
   max = INT_MIN;
   

3. Modify your loop such that the flag serves a purpose:

   Note:

   Don't waste the zeroth index of your array without any reason.
   Modify your array declaration and input loop accordingly before
   changing this one.

   for (i = 0; i < ARRAY_SIZE; ++i)
   {
    if (A[i] % 2 == 0)
    {
     foundAnswer = 1;
     if (A[i] < min) min = A[i];
     if (A[i] > max) max = A[i];
    }
   }
   

4. Modify your printing code slightly so that all the cases are
   covered.

   if (foundAnswer)
   {
    // Print min
    // Print max
   }
   else
   {
    // Print "min and max not found"
   }
   

Bonus:

You can learn from the following links in order to optimize your code:

1. How do I check if an integer is even or odd using bitwise operators

2. The algorithms discussed here will provide you with a better time complexity to achieve the same.

How do you find the max even sum from an array with using a number only once?

Appreciate that you will do one of two things with the total sum of all elements in the array. If the total sum be even, you return that number, but if it be odd, then you will have to remove a number in order to make it even. Note that removing an even number will leave the remaining sum odd, so there is no point in removing an even number. But, since the total sum be odd, then there must be at least one odd number in the array.

So all you need to do if the total sum be odd is to iterate over the array and remove the smallest odd number. And there is guaranteed to be at least one odd number.

Here is an implementation:

public int findMaxEvenSum(int[] array) {
    int total = 0;

    for (int i=0; i < array.length; ++i) {
        total += array[i];
    }

    if (total % 2 == 0) {
        return total;
    }

    // otherwise iterate over the array and remove the smallest odd
    // number from the sum
    int lastOdd = 0;

    for (int i=0; i < array.length; ++i) {
        if (array[i] % 2 == 1 && (lastOdd == 0 || array[i] < lastOdd)) {
            total += lastOdd;
            total -= array[i];
            lastOdd = array[i];
        }
    }

    return total;
}

How to find 3 largest even numbers in an array with C?

There are a few problems here:

1. You should only examine the array entries that are defined. Instead, you are looking at the entire array, including the undefined portion from nNumbers through MAX_NUMBERS-1. You will likely pick up garbage values there. Change your for loops to:

   for (i = 0; i < nNumbers; i++)
   

2. You are initializing greatest1, etc. to the first number in the array. That doesn't work if the number is odd and happens to be large enough to block the even number you're looking for.

3. If one of the largest even numbers occurs more than once, you will ignore the duplicates. For instance, if the largest number is 1000, and it occurs three times, you probably want to add all three and return 3000. You can fix this by keeping track of the indices you have chosen, and only rejecting a duplicate if the index matches, rather than the value.

Find the highest odd number in an array using recursion in java

While replacing the value of high with arr[index] if arr[index] value is higher just check if the value is odd. Else keep the high value as it is. Try the following code. It will return -1 if there have no odd value in the array(assuming all the values of array are positive, if there have possibility of negative value then you can replace -1 with integer min). Let me know if you don't understand any part.

public static int highestOdd(int [] arr, int index) {
   if(arr.length - 1 == index) return arr[index] % 2 > 0 ? arr[index] : -1;
   int high = highestOdd(arr,index+1);
   if(arr[index] % 2 > 0) high = high > arr[index] ? high : arr[index];
   return high;
}

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