Sum of Odd Numbers in an Array (Recursive)¶

Description¶

Write a recursive function that takes an array of numbers and returns the sum of only the odd numbers within that array.

Approach 1: Recursion from End¶

• Time Complexity: O(n) - The function calls itself for each element in the array.
• Space Complexity: O(n) - Each function call adds a frame to the call stack.

This function recursively calculates the sum by starting from the end of the array. - Base Case: When the index reaches 0, it checks if the first element is odd and returns it, otherwise returns 0. This stops the recursion. - Recursive Step: At the current index, it checks if the number is odd. - If it is odd, it adds the number to the result of the recursive call for the rest of the array (index - 1). - If it is even, it adds 0 to the result of the recursive call.

function sumOfOddNumbers(numbers: number[], index: number): number {
  // Base Case: Stop when we've processed the first element.
  if (index === 0) {
    return numbers[0] % 2 !== 0 ? numbers[0] : 0;
  }

  const isOdd = numbers[index] % 2 !== 0;

  // Recursive Step: Add the current number if it's odd, then process the rest of the array.
  return (isOdd ? numbers[index] : 0) + sumOfOddNumbers(numbers, index - 1);
}

const test = [10, 2, 5, 10, 3];
console.log(sumOfOddNumbers(test, test.length - 1)); // 8

const test2 = [1, 2, 3, 4, 5];
console.log(sumOfOddNumbers(test2, test2.length - 1)); // 9