Efficiently Summing Array Elements with Recursive Reduction

Question

Efficiently Summing Array Elements with Recursive Reduction

Find the sum of array element pairs by recursively consolidating them until only a single index remains.

EXAMPLES:

[1, 2, 3, 4, 5] => 48

Explanation:

The next array would be [3, 5, 7, 9] because [1+2, 2+3, 3+4, 4+5]
The next array would be [8, 12, 16] because [3+5, 5+7, 7+9]
The next array would be [20, 28] because [8+12, 12+16]
The final answer would be [48] because [20+28] and there are no more operands to add

I have implemented a solution for this problem using recursion, but I feel like there might be an easier approach. I'm trying to grasp the concept of recursion and how we determine the arguments passed into recursive calls. Can someone explain why we use (n - 1) or (n + 1) in our recursive calls and how we decide which one to use? I am also confused about passing arguments when returning from helper functions.

function reduceSum(input) {
  function simplify(input, index) {
    if (index === 1) {
      return input[0];
    }

    if (index === 0) {
      return 0;
    }

    for (let i = 0; i < input.length; i++) {
      input[i] += input[i + 1];
    }

    return simplify(input, index - 1);
  }

  return simplify(input, input.length);
}


console.log(reduceSum([1, 2, 3, 4]) == 20)
console.log(reduceSum([5]) == 5)
console.log(reduceSum([]) == 0)
console.log(reduceSum([1, 3, 5]) == 12)
console.log(reduceSum([-5, 5]) == 0)
console.log(reduceSum([-5, 5, -5, 5]) == 0)
console.log(reduceSum([-5, 5, 5, -5]) == 20)

javascript arrays recursion sum reduce

Answer 1

Answer №1

Here's an alternative approach that may not necessarily simplify the solution, as it essentially accomplishes the same task (without altering the original array). However, there could still be some valuable insights to be gained from this method:

function reduceSum(input) {
  if(input.length <= 1)
    return input[0] || 0;
  
  return reduceSum(
    input.reduce(
      (acc, val, i, { [i + 1]: next }) =>
        typeof next !== 'undefined' ? [ ...acc, val + next ] : acc,
      []
    )
  );
}


console.log(reduceSum([1, 2, 3, 4]) == 20)
console.log(reduceSum([5]) == 5)
console.log(reduceSum([]) == 0)
console.log(reduceSum([1, 3, 5]) == 12)
console.log(reduceSum([-5, 5]) == 0)
console.log(reduceSum([-5, 5, -5, 5]) == 0)
console.log(reduceSum([-5, 5, 5, -5]) == 20)

The next variable is assigned using object destructuring within the array argument passed to the reduce callback function. When the last element of the array is processed by the reduce method, `next` becomes undefined and gets skipped. This leads to a reduction in the size of the array with each execution of reduceSum, similar to the original behavior you described.

Answer 2

Here's an alternative approach that may not necessarily simplify the solution, as it essentially accomplishes the same task (without altering the original array). However, there could still be some valuable insights to be gained from this method:

function reduceSum(input) {
  if(input.length <= 1)
    return input[0] || 0;
  
  return reduceSum(
    input.reduce(
      (acc, val, i, { [i + 1]: next }) =>
        typeof next !== 'undefined' ? [ ...acc, val + next ] : acc,
      []
    )
  );
}


console.log(reduceSum([1, 2, 3, 4]) == 20)
console.log(reduceSum([5]) == 5)
console.log(reduceSum([]) == 0)
console.log(reduceSum([1, 3, 5]) == 12)
console.log(reduceSum([-5, 5]) == 0)
console.log(reduceSum([-5, 5, -5, 5]) == 0)
console.log(reduceSum([-5, 5, 5, -5]) == 20)

The next variable is assigned using object destructuring within the array argument passed to the reduce callback function. When the last element of the array is processed by the reduce method, `next` becomes undefined and gets skipped. This leads to a reduction in the size of the array with each execution of reduceSum, similar to the original behavior you described.

Answer 3

Answer №2

Every iteration within the loop modifies the elements in the array. To prevent an infinite loop, the index must be decremented after each iteration. To observe the progression of the 'input' variable with each call to the 'simplify()' function, simply include the statement

console.log("" + input)

immediately after the 'for' loop.

There are several ways to optimize your code:

Avoid using the 'simplify()' function altogether and instead check if the 'index' variable is 'undefined':

function reduceSum(input, index) {
  if (index === undefined)
    index = input.length;

  if (index === 1) {
    return input[0];
  }

  if (index === 0) {
    return 0;
  }

  for (let i = 0; i < input.length; i++) {
    input[i] += input[i + 1];
  }
  console.log("input: " + input);
  return reduceSum(input, index - 1);
}

console.log("result:", reduceSum([1, 2, 3, 4, 5]) == 48)
console.log("result:", reduceSum([1, 2, 3, 4]) == 20)
console.log("result:", reduceSum([5]) == 5)
console.log("result:", reduceSum([]) == 0)
console.log("result:", reduceSum([1, 3, 5]) == 12)
console.log("result:", reduceSum([-5, 5]) == 0)
console.log("result:", reduceSum([-5, 5, -5, 5]) == 0)
console.log("result:", reduceSum([-5, 5, 5, -5]) == 20)

You can also improve efficiency by iterating only from 0 to 'index - 1' instead of covering the entire array:

function reduceSum(input, index) {
  if (index === undefined)
    index = input.length;

  if (index === 1) {
    return input[0];
  }

  if (index === 0) {
    return 0;
  }

  for (let i = 0; i < index - 1; i++) {
    input[i] += input[i + 1];
  }
  console.log("input: " + input);
  return reduceSum(input, index - 1);
}

console.log("result:", reduceSum([1, 2, 3, 4, 5]) == 48)
console.log("result:", reduceSum([1, 2, 3, 4]) == 20)
console.log("result:", reduceSum([5]) == 5)
console.log("result:", reduceSum([]) == 0)
console.log("result:", reduceSum([1, 3, 5]) == 12)
console.log("result:", reduceSum([-5, 5]) == 0)
console.log("result:", reduceSum([-5, 5, -5, 5]) == 0)
console.log("result:", reduceSum([-5, 5, 5, -5]) == 20)

For further optimization, provide a default value for the 'index' variable and move conditions to the end of the function. This eliminates the need for an additional function call to obtain the final result:

function reduceSum(input, index = input.length - 1) {
  for (let i = 0; i < index; i++) {
    input[i] += input[i + 1];
  }
  console.log("input: " + input);
  if (index > 1)
    return reduceSum(input, --index);

  //return the final result, or 0 if the array is empty
  return input[0] || 0;
}

console.log("result:", reduceSum([1, 2, 3, 4, 5]) == 48)
console.log("result:", reduceSum([1, 2, 3, 4]) == 20)
console.log("result:", reduceSum([5]) == 5)
console.log("result:", reduceSum([]) == 0)
console.log("result:", reduceSum([1, 3, 5]) == 12)
console.log("result:", reduceSum([-5, 5]) == 0)
console.log("result:", reduceSum([-5, 5, -5, 5]) == 0)
console.log("result:", reduceSum([-5, 5, 5, -5]) == 20)

Finally, you can eliminate the use of the 'index' variable entirely by removing the last item from the 'input' array. Note that this method may be slower than the previous ones:

function reduceSum(input) {
  for (let i = 0; i < input.length - 1; i++) {
    input[i] += input[i + 1];
  }
  console.log("input: " + input);
  if (input.length > 1)
  {
    //remove the last item by reducing the length of the array
    input.length--;
    return reduceSum(input);
  }
  //return the final result, or 0 if the array is empty
  return input[0] || 0;
}

console.log("result:", reduceSum([1, 2, 3, 4, 5]) == 48)
console.log("result:", reduceSum([1, 2, 3, 4]) == 20)
console.log("result:", reduceSum([5]) == 5)
console.log("result:", reduceSum([]) == 0)
console.log("result:", reduceSum([1, 3, 5]) == 12)
console.log("result:", reduceSum([-5, 5]) == 0)
console.log("result:", reduceSum([-5, 5, -5, 5]) == 0)
console.log("result:", reduceSum([-5, 5, 5, -5]) == 20)

Answer 4

Every iteration within the loop modifies the elements in the array. To prevent an infinite loop, the index must be decremented after each iteration. To observe the progression of the 'input' variable with each call to the 'simplify()' function, simply include the statement

console.log("" + input)

immediately after the 'for' loop.

There are several ways to optimize your code:

Avoid using the 'simplify()' function altogether and instead check if the 'index' variable is 'undefined':

function reduceSum(input, index) {
  if (index === undefined)
    index = input.length;

  if (index === 1) {
    return input[0];
  }

  if (index === 0) {
    return 0;
  }

  for (let i = 0; i < input.length; i++) {
    input[i] += input[i + 1];
  }
  console.log("input: " + input);
  return reduceSum(input, index - 1);
}

console.log("result:", reduceSum([1, 2, 3, 4, 5]) == 48)
console.log("result:", reduceSum([1, 2, 3, 4]) == 20)
console.log("result:", reduceSum([5]) == 5)
console.log("result:", reduceSum([]) == 0)
console.log("result:", reduceSum([1, 3, 5]) == 12)
console.log("result:", reduceSum([-5, 5]) == 0)
console.log("result:", reduceSum([-5, 5, -5, 5]) == 0)
console.log("result:", reduceSum([-5, 5, 5, -5]) == 20)

You can also improve efficiency by iterating only from 0 to 'index - 1' instead of covering the entire array:

function reduceSum(input, index) {
  if (index === undefined)
    index = input.length;

  if (index === 1) {
    return input[0];
  }

  if (index === 0) {
    return 0;
  }

  for (let i = 0; i < index - 1; i++) {
    input[i] += input[i + 1];
  }
  console.log("input: " + input);
  return reduceSum(input, index - 1);
}

console.log("result:", reduceSum([1, 2, 3, 4, 5]) == 48)
console.log("result:", reduceSum([1, 2, 3, 4]) == 20)
console.log("result:", reduceSum([5]) == 5)
console.log("result:", reduceSum([]) == 0)
console.log("result:", reduceSum([1, 3, 5]) == 12)
console.log("result:", reduceSum([-5, 5]) == 0)
console.log("result:", reduceSum([-5, 5, -5, 5]) == 0)
console.log("result:", reduceSum([-5, 5, 5, -5]) == 20)

For further optimization, provide a default value for the 'index' variable and move conditions to the end of the function. This eliminates the need for an additional function call to obtain the final result:

function reduceSum(input, index = input.length - 1) {
  for (let i = 0; i < index; i++) {
    input[i] += input[i + 1];
  }
  console.log("input: " + input);
  if (index > 1)
    return reduceSum(input, --index);

  //return the final result, or 0 if the array is empty
  return input[0] || 0;
}

console.log("result:", reduceSum([1, 2, 3, 4, 5]) == 48)
console.log("result:", reduceSum([1, 2, 3, 4]) == 20)
console.log("result:", reduceSum([5]) == 5)
console.log("result:", reduceSum([]) == 0)
console.log("result:", reduceSum([1, 3, 5]) == 12)
console.log("result:", reduceSum([-5, 5]) == 0)
console.log("result:", reduceSum([-5, 5, -5, 5]) == 0)
console.log("result:", reduceSum([-5, 5, 5, -5]) == 20)

Finally, you can eliminate the use of the 'index' variable entirely by removing the last item from the 'input' array. Note that this method may be slower than the previous ones:

function reduceSum(input) {
  for (let i = 0; i < input.length - 1; i++) {
    input[i] += input[i + 1];
  }
  console.log("input: " + input);
  if (input.length > 1)
  {
    //remove the last item by reducing the length of the array
    input.length--;
    return reduceSum(input);
  }
  //return the final result, or 0 if the array is empty
  return input[0] || 0;
}

console.log("result:", reduceSum([1, 2, 3, 4, 5]) == 48)
console.log("result:", reduceSum([1, 2, 3, 4]) == 20)
console.log("result:", reduceSum([5]) == 5)
console.log("result:", reduceSum([]) == 0)
console.log("result:", reduceSum([1, 3, 5]) == 12)
console.log("result:", reduceSum([-5, 5]) == 0)
console.log("result:", reduceSum([-5, 5, -5, 5]) == 0)
console.log("result:", reduceSum([-5, 5, 5, -5]) == 20)

Answer 5

Answer №3

If you want to implement a for loop, you can do so by following this example:

function calculateSum(arr) {
  if (arr.length < 2) {
    return arr[0] || 0;
  }
  const result = [];
  for(let i = 0; i < arr.length - 1; i++) {
    result.push(arr[i] + arr[i+1]);
  }
  return calculateSum(result);
}

console.log(calculateSum([1, 2, 3, 4]) == 20)
console.log(calculateSum([5]) == 5)
console.log(calculateSum([]) == 0)
console.log(calculateSum([1, 3, 5]) == 12)
console.log(calculateSum([-5, 5]) == 0)
console.log(calculateSum([-5, 5, -5, 5]) == 0)
console.log(calculateSum([-5, 5, 5, -5]) == 20)

Answer 6

If you want to implement a for loop, you can do so by following this example:

function calculateSum(arr) {
  if (arr.length < 2) {
    return arr[0] || 0;
  }
  const result = [];
  for(let i = 0; i < arr.length - 1; i++) {
    result.push(arr[i] + arr[i+1]);
  }
  return calculateSum(result);
}

console.log(calculateSum([1, 2, 3, 4]) == 20)
console.log(calculateSum([5]) == 5)
console.log(calculateSum([]) == 0)
console.log(calculateSum([1, 3, 5]) == 12)
console.log(calculateSum([-5, 5]) == 0)
console.log(calculateSum([-5, 5, -5, 5]) == 0)
console.log(calculateSum([-5, 5, 5, -5]) == 20)

Efficiently Summing Array Elements with Recursive Reduction

Answer №1

Answer №2

Answer №3

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