Discover a series of sequential numbers that result from being reduced by 1 using JavaScript

Question

Discover a series of sequential numbers that result from being reduced by 1 using JavaScript

Trying to identify consecutive numbers from an array of numbers has been my latest coding challenge. Here's the snippet I've come up with:

let arr = [4, 3, 5, 4, 3];
let new_arr = [];
for (let i = 0; i < arr.length; i++) {
    let internalArr = [arr[i]];
    new_arr.push(internalArr);
    if (arr[i] - arr[i + 1] === 1) {
        new_arr.push([arr[i], arr[i + 1]]);
    }
}
console.log(new_arr);

I'm struggling with getting more than two elements into the new array. For instance, [5, 4, 3] should also be included in the new_arr. However, currently, I'm only able to form arrays with two elements. Any insights on what might be missing?

javascript arrays sequence push subsequence

Answer 1

Answer №1

To achieve the desired outcome, it is recommended to implement a nested loop structure. One way to do this is by iterating through each index starting from i, and within an inner loop, checking if the subsequent number is 1 less than the current one. If this condition is met, a new array can be created and added to the collection - otherwise, the inner loop breaks out and moves on to the next iteration of the outer loop.

let inputArray = [4, 3, 5, 4, 3];
const allSequences = [];
for (let i = 0; i < inputArray.length; i++) {
  let oneSequence = [inputArray[i]];
  allSequences.push(oneSequence);
  for (let j = i + 1; j < inputArray.length; j++) {
    if (inputArray[j] === oneSequence[oneSequence.length - 1] - 1) {
      oneSequence = [...oneSequence, inputArray[j]];
      allSequences.push(oneSequence);
    } else {
      break;
    }
  }
}
console.log(allSequences);

Answer 2

To achieve the desired outcome, it is recommended to implement a nested loop structure. One way to do this is by iterating through each index starting from i, and within an inner loop, checking if the subsequent number is 1 less than the current one. If this condition is met, a new array can be created and added to the collection - otherwise, the inner loop breaks out and moves on to the next iteration of the outer loop.

let inputArray = [4, 3, 5, 4, 3];
const allSequences = [];
for (let i = 0; i < inputArray.length; i++) {
  let oneSequence = [inputArray[i]];
  allSequences.push(oneSequence);
  for (let j = i + 1; j < inputArray.length; j++) {
    if (inputArray[j] === oneSequence[oneSequence.length - 1] - 1) {
      oneSequence = [...oneSequence, inputArray[j]];
      allSequences.push(oneSequence);
    } else {
      break;
    }
  }
}
console.log(allSequences);

Answer 3

Answer №2

To create groups of consecutive numbers, you can organize them into sub arrays by following the guidelines in the code snippet below:

// List of numbers
const arr = [4,3,5,4,3,7,2];

// Initialize a new array
const new_arr = [];

// Array to hold consecutive numbers
let sub_arr = [];

// Iterate through the numbers
for(let i=0; i<arr.length; i++) {
  // Add current number to sub array
  sub_arr.push(arr[i]);
  
  // Check if next number is not consecutive
  // If so, add sub array to new array and reset sub array
  if(arr[i] - arr[i+1] != 1) {
    new_arr.push(sub_arr);
    sub_arr = [];
  }
}

// Display result
console.log(new_arr);

If you only want consecutive groups with more than one element:

// List of numbers
const arr = [4,3,5,4,3,7,2];

// Initialize a new array
const new_arr = [];

// Array to hold consecutive numbers
let sub_arr = [];

// Iterate through the numbers
for(let i=0; i<arr.length; i++) {
  // Add current number to sub array
  sub_arr.push(arr[i]);
  
  // Check if next number is not consecutive
  // If so, add sub array to new array and reset sub array
  if(arr[i] - arr[i+1] != 1) {
    // Only add sub array to new array 
    // if it contains more than 1 element
    if(sub_arr.length > 1) new_arr.push(sub_arr);
    sub_arr = [];
  }
}

// Display result
console.log(new_arr);

Answer 4

To create groups of consecutive numbers, you can organize them into sub arrays by following the guidelines in the code snippet below:

// List of numbers
const arr = [4,3,5,4,3,7,2];

// Initialize a new array
const new_arr = [];

// Array to hold consecutive numbers
let sub_arr = [];

// Iterate through the numbers
for(let i=0; i<arr.length; i++) {
  // Add current number to sub array
  sub_arr.push(arr[i]);
  
  // Check if next number is not consecutive
  // If so, add sub array to new array and reset sub array
  if(arr[i] - arr[i+1] != 1) {
    new_arr.push(sub_arr);
    sub_arr = [];
  }
}

// Display result
console.log(new_arr);

If you only want consecutive groups with more than one element:

// List of numbers
const arr = [4,3,5,4,3,7,2];

// Initialize a new array
const new_arr = [];

// Array to hold consecutive numbers
let sub_arr = [];

// Iterate through the numbers
for(let i=0; i<arr.length; i++) {
  // Add current number to sub array
  sub_arr.push(arr[i]);
  
  // Check if next number is not consecutive
  // If so, add sub array to new array and reset sub array
  if(arr[i] - arr[i+1] != 1) {
    // Only add sub array to new array 
    // if it contains more than 1 element
    if(sub_arr.length > 1) new_arr.push(sub_arr);
    sub_arr = [];
  }
}

// Display result
console.log(new_arr);

Discover a series of sequential numbers that result from being reduced by 1 using JavaScript

Answer №1

Answer №2

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