Identify and remove numbers from an array that have the same digit, without having prior knowledge of which digit it

Question

Identify and remove numbers from an array that have the same digit, without having prior knowledge of which digit it

Is there an easy way to separate values in an array of random numbers between 0-99 based on their first and second digits? I need two arrays: one for values that share the same first digit, and another for values that share the same second digit. The actual digit value doesn't matter; as long as it is repeated within a number. Just to clarify, the digits 0-9 should be treated as if the first digit was 0.

For example, given the array:

[0, 10, 20, 11, 19, 12, 54, 64, 23, 24]

, the resulting arrays would be: [10, 11, 12, 19, 23, 24] and [0, 10, 20, 24, 54, 64]. Some values can appear in both arrays if they meet the specified criteria.

I have come across solutions that work when you know the specific digit you are looking for, but in this case, I am unsure how to proceed.

One suggested approach is to convert the numbers in the array into strings using array.map(String), so that the first and second digits can be accessed as first[0] and second[1]. However, I am unsure about the next steps. Any suggestions would be greatly appreciated.

javascript arrays

Answer 1

Answer №1

To organize your array by grouping numbers based on their first and last digits, create an object accumulator. Then, sort the grouped array based on their length.

const input = [0, 10, 20, 11, 19, 12, 54, 64, 23, 24],
      groupLastDigits = input.reduce((result, num) => {
        const lastDigit = num % 10;
        result[lastDigit] ??= [];
        result[lastDigit].push(num);
        return result;
      }, {}),
      groupFirstDigits = input.reduce((result, num) => {
        let firstDigit = Math.floor(num / 10);
        if(firstDigit === num) firstDigit = 0;
        result[firstDigit] ??= [];
        result[firstDigit].push(num);
        return result;
      }, {}),
      extractNumbers = obj => Object.values(obj).filter(arr => arr.length > 1).flat(),
      commonLastDigits = extractNumbers(groupLastDigits),
      commonFirstDigits = extractNumbers(groupFirstDigits);
console.log(commonLastDigits);
console.log(commonFirstDigits);

Answer 2

To organize your array by grouping numbers based on their first and last digits, create an object accumulator. Then, sort the grouped array based on their length.

const input = [0, 10, 20, 11, 19, 12, 54, 64, 23, 24],
      groupLastDigits = input.reduce((result, num) => {
        const lastDigit = num % 10;
        result[lastDigit] ??= [];
        result[lastDigit].push(num);
        return result;
      }, {}),
      groupFirstDigits = input.reduce((result, num) => {
        let firstDigit = Math.floor(num / 10);
        if(firstDigit === num) firstDigit = 0;
        result[firstDigit] ??= [];
        result[firstDigit].push(num);
        return result;
      }, {}),
      extractNumbers = obj => Object.values(obj).filter(arr => arr.length > 1).flat(),
      commonLastDigits = extractNumbers(groupLastDigits),
      commonFirstDigits = extractNumbers(groupFirstDigits);
console.log(commonLastDigits);
console.log(commonFirstDigits);

Answer 3

Answer №2

Presenting a versatile solution for dealing with n digits!

function* identifyMatchingDigitClusters(inputArray) {
  
  // Keep in mind that very large numbers may be converted to strings
  // in scientific notation, potentially affecting the approach
  let maxDigits = Math.max(...inputArray.map(number => `${number}`.length));
  let stringsArr = inputArray.map(number => {
    let str = `${number}`;
    while (str.length < maxDigits) str = `0${str}`;
    return str;
  });
  
  for (let i = 0; i < maxDigits; i++) {
    
    let result = [];
    for (let n = 0; n <= 9; n++) {
      let matchingDigits = stringsArr.filter(str => str[i] === `${n}`);
      if (matchingDigits.length > 1) result.push(...matchingDigits);
    }
    yield result;
    
  }
  
};

let testCases = [
  [ 0, 10, 20, 11, 19, 12, 54, 64, 23, 24 ],
  [ 100, 200, 300 ],
  [ 111, 222, 333, 771, 828, 399 ],
  [],
  [ 1 ],
  [ 92929, 91919 ]
];
for (let inputArr of testCases) {
  console.log(`Digit Clusters for: [ ${inputArr.join(', ')} ]`);
  for (let cluster of identifyMatchingDigitClusters(inputArr)) console.log(`  [ ${cluster.join(', ')} ]`);
}

Answer 4

Presenting a versatile solution for dealing with n digits!

function* identifyMatchingDigitClusters(inputArray) {
  
  // Keep in mind that very large numbers may be converted to strings
  // in scientific notation, potentially affecting the approach
  let maxDigits = Math.max(...inputArray.map(number => `${number}`.length));
  let stringsArr = inputArray.map(number => {
    let str = `${number}`;
    while (str.length < maxDigits) str = `0${str}`;
    return str;
  });
  
  for (let i = 0; i < maxDigits; i++) {
    
    let result = [];
    for (let n = 0; n <= 9; n++) {
      let matchingDigits = stringsArr.filter(str => str[i] === `${n}`);
      if (matchingDigits.length > 1) result.push(...matchingDigits);
    }
    yield result;
    
  }
  
};

let testCases = [
  [ 0, 10, 20, 11, 19, 12, 54, 64, 23, 24 ],
  [ 100, 200, 300 ],
  [ 111, 222, 333, 771, 828, 399 ],
  [],
  [ 1 ],
  [ 92929, 91919 ]
];
for (let inputArr of testCases) {
  console.log(`Digit Clusters for: [ ${inputArr.join(', ')} ]`);
  for (let cluster of identifyMatchingDigitClusters(inputArr)) console.log(`  [ ${cluster.join(', ')} ]`);
}

Identify and remove numbers from an array that have the same digit, without having prior knowledge of which digit it

Answer №1

Answer №2

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