Carrying over additions in arrays using only Javascript

Question

Carrying over additions in arrays using only Javascript

I'd like to implement a basic array addition with carryover. Additionally, I need to display the carryover and result values.

For instance:

e.g

var input = [[0,0,9],[0,9,9]];
 var carryover = [];
var result = [];

Thank you!

javascript arrays math addition

Answer 1

Answer №1

If you've been struggling with two parts, it's likely how to calculate the carry and pinpointing the result accurately.

result [digit] = t % 10;

The % 10 operation is known as modulus. In this case, we are applying a modulus by 10 to determine the value of the tens place.

carryover [digit] = Math.trunc(t / 10);

To find the carryover, you simply divide by 10 and then utilize Math.trunc to remove any decimal points.

var input = [[0,0,0,9],[0,9,9]];
var carryover = [];
var result = [];

var digitLength = Math.max(input[0].length, input[1].length);

// Standardizing inputs to have equal lengths
input[0].unshift(
  ...new Array(digitLength - input[0].length).fill(0));
input[1].unshift(
  ...new Array(digitLength - input[1].length).fill(0));

for (var digit = digitLength - 1; digit >= 0; digit -= 1) {
  var t = input[0][digit] + input[1][digit];
  if (digit < digitLength - 1) 
    t += carryover[digit + 1];
  result [digit] = t % 10;
  carryover [digit] = Math.trunc(t / 10);
}
result.unshift(carryover[0]);

console.log('result: ' + result.join(', '));
console.log('carry:  ' + carryover.join(', '));

Answer 2

If you've been struggling with two parts, it's likely how to calculate the carry and pinpointing the result accurately.

result [digit] = t % 10;

The % 10 operation is known as modulus. In this case, we are applying a modulus by 10 to determine the value of the tens place.

carryover [digit] = Math.trunc(t / 10);

To find the carryover, you simply divide by 10 and then utilize Math.trunc to remove any decimal points.

var input = [[0,0,0,9],[0,9,9]];
var carryover = [];
var result = [];

var digitLength = Math.max(input[0].length, input[1].length);

// Standardizing inputs to have equal lengths
input[0].unshift(
  ...new Array(digitLength - input[0].length).fill(0));
input[1].unshift(
  ...new Array(digitLength - input[1].length).fill(0));

for (var digit = digitLength - 1; digit >= 0; digit -= 1) {
  var t = input[0][digit] + input[1][digit];
  if (digit < digitLength - 1) 
    t += carryover[digit + 1];
  result [digit] = t % 10;
  carryover [digit] = Math.trunc(t / 10);
}
result.unshift(carryover[0]);

console.log('result: ' + result.join(', '));
console.log('carry:  ' + carryover.join(', '));

Answer 3

Answer №2

    1.convert both numbers into arrays of individual digits and reverse the order.
    2.calculate the end index for the for-loop based on the maximum length of the two arrays.
    3.initialize a third array to store carryover digits with zeros (remember the extra digit).
    4.Add the corresponding digits from step 1 and step 3,
       while iterating through each digit from right to left,
    4.1 if the sum is greater than 9, add 1 to the next carryover position.
    5. at the end of the loop, you will have an array of carried over digits,
         count the number of 1s in this array.

function numberOfCarryOperations(number1, number2) {
    const digits1 = [...number1.toString()].reverse()
    const digits2 = [...number2.toString()].reverse()
    const endIndex = Math.max(digits1.length, digits2.length)
    const carryOverArray = Array(endIndex + 1).fill(0)
    for (let j = 0; j < endIndex; j++) {
        if (((Number(digits1[j] ? digits1[j] : 0)) + Number(digits2[j] ? digits2[j] : 0) + carryOverArray[j]) > 9) {
            carryOverArray[j + 1] = 1
        }
    }
    return carryOverArray.reduce((total, current) => total + current)
}

Answer 4

    1.convert both numbers into arrays of individual digits and reverse the order.
    2.calculate the end index for the for-loop based on the maximum length of the two arrays.
    3.initialize a third array to store carryover digits with zeros (remember the extra digit).
    4.Add the corresponding digits from step 1 and step 3,
       while iterating through each digit from right to left,
    4.1 if the sum is greater than 9, add 1 to the next carryover position.
    5. at the end of the loop, you will have an array of carried over digits,
         count the number of 1s in this array.

function numberOfCarryOperations(number1, number2) {
    const digits1 = [...number1.toString()].reverse()
    const digits2 = [...number2.toString()].reverse()
    const endIndex = Math.max(digits1.length, digits2.length)
    const carryOverArray = Array(endIndex + 1).fill(0)
    for (let j = 0; j < endIndex; j++) {
        if (((Number(digits1[j] ? digits1[j] : 0)) + Number(digits2[j] ? digits2[j] : 0) + carryOverArray[j]) > 9) {
            carryOverArray[j + 1] = 1
        }
    }
    return carryOverArray.reduce((total, current) => total + current)
}

Answer 5

Answer №3

My solution to the problem is designed to handle multiple input arrays with varying lengths:

The number of items in each array can differ
All types of input arrays are accepted

I have included explanatory comments within the code to make it easier to understand the logic behind the solution.

const
  input = [
    [0,0,9],
    [0,9,9],
    [1,0,9,9]
  ];
  
function getMaxArrayLength(values) {
  // This function determines the length of the longest array by using reduce with an initial value of 0.
  return values.reduce((maxLength, array) => {
      // Compare and return the larger value between the previous length and the current array's length.
      return Math.max(maxLength, array.length);
    }, 0);
}
  
function sumValues(values) {
  const
    maxLength = getMaxArrayLength(values),
    result = [],
    carry = [];

  for (let index = 1; index <= maxLength; index++) {
    const
      carryValue = (carry.length === 0) ? 0 : carry[carry.length-1],
      
      sum = values.reduce((sum, array) => { 
        const 
          arrayIndex = array.length - index;
        
        if (arrayIndex < 0) {
          return sum;
        }
        
        return sum + array[arrayIndex];
      }, 0) + carryValue;
     
    carry.push(Math.floor(sum / 10));
    
    result.push(sum % 10);
  }
  
  return {
    carryOver: carry.reverse(),
    result: result.reverse()
  };
}

const
  result = sumValues(input);
  

console.log(`Carry over: ${result.carryOver}`);
console.log(`Result: ${result.result}`);

Answer 6

My solution to the problem is designed to handle multiple input arrays with varying lengths:

The number of items in each array can differ
All types of input arrays are accepted

I have included explanatory comments within the code to make it easier to understand the logic behind the solution.

const
  input = [
    [0,0,9],
    [0,9,9],
    [1,0,9,9]
  ];
  
function getMaxArrayLength(values) {
  // This function determines the length of the longest array by using reduce with an initial value of 0.
  return values.reduce((maxLength, array) => {
      // Compare and return the larger value between the previous length and the current array's length.
      return Math.max(maxLength, array.length);
    }, 0);
}
  
function sumValues(values) {
  const
    maxLength = getMaxArrayLength(values),
    result = [],
    carry = [];

  for (let index = 1; index <= maxLength; index++) {
    const
      carryValue = (carry.length === 0) ? 0 : carry[carry.length-1],
      
      sum = values.reduce((sum, array) => { 
        const 
          arrayIndex = array.length - index;
        
        if (arrayIndex < 0) {
          return sum;
        }
        
        return sum + array[arrayIndex];
      }, 0) + carryValue;
     
    carry.push(Math.floor(sum / 10));
    
    result.push(sum % 10);
  }
  
  return {
    carryOver: carry.reverse(),
    result: result.reverse()
  };
}

const
  result = sumValues(input);
  

console.log(`Carry over: ${result.carryOver}`);
console.log(`Result: ${result.result}`);

Carrying over additions in arrays using only Javascript

Answer №1

Answer №2

Answer №3

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