Determining the largest range possible in a sorted array of integers

Question

Determining the largest range possible in a sorted array of integers

I need help with a JavaScript implementation for the following challenge.
Imagine we have a sorted array:

[1,2,5,9,10,12,20,21,22,23,24,26,27]

I want to find the length of the longest consecutive range that increments by 1 without duplicates.

In the provided example, we have the following ranges:

1,2
9,10
20,21,22,23,24 // longest range here
26,27

Therefore, the expected result for this example should be 5.

While I know how to solve this problem using a straightforward method, I am curious if there is a more efficient and concise algorithm available.

javascript arrays algorithm sorting

Answer 1

Answer №1

An Alternative Approach

While my solution may not be more efficient than others, it offers a concise code that iterates over the array only once, except for the initial element. Below is my proposed solution:

var arr = [1, 2, 5, 9, 10, 12, 20, 21, 22, 23, 24, 26, 27];
var streak = 0, best = 0, bestStart;
for (var i = 1; i < arr.length; i++) {
  if(arr[i]-arr[i-1] === 1) streak++;
  else streak = 0;
  if (streak > best) [best, bestStart] = [streak, i - streak];
}
var bestArr = arr.slice(bestStart, bestStart + best + 1);
console.log('Best streak: '+bestArr);

Optimizing Performance

Upon reviewing the code, I identified a way to enhance its performance slightly by avoiding checking the last few elements of the array based on the previous value of best:

var arr = [1, 2, 5, 9, 10, 12, 20, 21, 22, 23, 24, 26, 27];
var streak = 0, best = 0, bestStart;
for (var i = 1; i < arr.length; i++) {
  if(best > arr.length - i + streak) break;
  if(arr[i]-arr[i-1] === 1) streak++;
  else streak = 0;
  if (streak > best) [best, bestStart] = [streak, i - streak];
}
var bestArr = arr.slice(bestStart, bestStart + best + 1);
console.log('Best streak: '+bestArr);

Answer 2

An Alternative Approach

While my solution may not be more efficient than others, it offers a concise code that iterates over the array only once, except for the initial element. Below is my proposed solution:

var arr = [1, 2, 5, 9, 10, 12, 20, 21, 22, 23, 24, 26, 27];
var streak = 0, best = 0, bestStart;
for (var i = 1; i < arr.length; i++) {
  if(arr[i]-arr[i-1] === 1) streak++;
  else streak = 0;
  if (streak > best) [best, bestStart] = [streak, i - streak];
}
var bestArr = arr.slice(bestStart, bestStart + best + 1);
console.log('Best streak: '+bestArr);

Optimizing Performance

Upon reviewing the code, I identified a way to enhance its performance slightly by avoiding checking the last few elements of the array based on the previous value of best:

var arr = [1, 2, 5, 9, 10, 12, 20, 21, 22, 23, 24, 26, 27];
var streak = 0, best = 0, bestStart;
for (var i = 1; i < arr.length; i++) {
  if(best > arr.length - i + streak) break;
  if(arr[i]-arr[i-1] === 1) streak++;
  else streak = 0;
  if (streak > best) [best, bestStart] = [streak, i - streak];
}
var bestArr = arr.slice(bestStart, bestStart + best + 1);
console.log('Best streak: '+bestArr);

Answer 3

Answer №2

To find a solution, we can iterate through the array and maintain the current range as long as the numbers are consecutive. When we encounter a number that is not a successor of the previous number, we close the current range and compare its length to the length of the last range.

This iterative approach allows us to update the maximum range length in constant time, resulting in an O(n) algorithm where n is the number of elements in the input array.

A pseudocode implementation similar to C# could look like this:

int MaximumLength = minus infinity;
int CurrentValue = Input[0];
int CurrentLength = 1;

for(int i = 1; i < Input.Length; i++)
{
    if (CurrentValue + 1 == Input[i])
    {
        // same range
        CurrentLength++;
    }
    else
    {
        // new range
        MaximumLength = Math.Max(MaximumLength, CurrentLength);
        CurrentLength = 1;
    }
    CurrentValue = Input[i];
}
// check current length after loop termination
MaximumLength = Math.Max(MaximumLength, CurrentLength);

The time complexity of this algorithm cannot be improved beyond O(n) because reading the input alone requires at least O(n) time. This implies that every element in the input affects the result, making it necessary to examine each one. Even Philipp Maurer's suggested algorithm would also have an O(n) runtime when the maximum range length is only 1, meaning there are no consecutive numbers in the input array.

Answer 4

To find a solution, we can iterate through the array and maintain the current range as long as the numbers are consecutive. When we encounter a number that is not a successor of the previous number, we close the current range and compare its length to the length of the last range.

This iterative approach allows us to update the maximum range length in constant time, resulting in an O(n) algorithm where n is the number of elements in the input array.

A pseudocode implementation similar to C# could look like this:

int MaximumLength = minus infinity;
int CurrentValue = Input[0];
int CurrentLength = 1;

for(int i = 1; i < Input.Length; i++)
{
    if (CurrentValue + 1 == Input[i])
    {
        // same range
        CurrentLength++;
    }
    else
    {
        // new range
        MaximumLength = Math.Max(MaximumLength, CurrentLength);
        CurrentLength = 1;
    }
    CurrentValue = Input[i];
}
// check current length after loop termination
MaximumLength = Math.Max(MaximumLength, CurrentLength);

The time complexity of this algorithm cannot be improved beyond O(n) because reading the input alone requires at least O(n) time. This implies that every element in the input affects the result, making it necessary to examine each one. Even Philipp Maurer's suggested algorithm would also have an O(n) runtime when the maximum range length is only 1, meaning there are no consecutive numbers in the input array.

Answer 5

Answer №3

To improve efficiency, it's recommended to calculate the maximum length first rather than last in this context.

Initialize max as 0
Set n as the length of the array
While n is greater than 2
  Initialize m as 0
  While m is less than or equal to (the length of the array - n)
    Set first as m
    Set last as m + n - 1 
    Calculate the difference as (value of element 'last' in array) - (value of element 'first' in array)
    if diff equals n - 1 then
      Update max to be n
      Stop the loop
    end if
    Increment m
  end while
  Decrement n
end while

Edit (javascript implementation)

var a = [1,2,5,9,10,12,20,21,22,23,24,26,27];
var max = 1;
var n = a.length;
while(n is greater than 2) {
  var m = 0;
  while(m is less than or equal to a.length - n)
  {
    var first = m;
    var last = m + n - 1;
    var diff = a[last] - a[first];
    if (diff is equal to n - 1 && diff is greater than max) {
      Update max to be n;
      Break out of inner loop;
    }
    increment m;
    }
  decrement n;
}

console.log(max);

JSFiddle

Answer 6

To improve efficiency, it's recommended to calculate the maximum length first rather than last in this context.

Initialize max as 0
Set n as the length of the array
While n is greater than 2
  Initialize m as 0
  While m is less than or equal to (the length of the array - n)
    Set first as m
    Set last as m + n - 1 
    Calculate the difference as (value of element 'last' in array) - (value of element 'first' in array)
    if diff equals n - 1 then
      Update max to be n
      Stop the loop
    end if
    Increment m
  end while
  Decrement n
end while

Edit (javascript implementation)

var a = [1,2,5,9,10,12,20,21,22,23,24,26,27];
var max = 1;
var n = a.length;
while(n is greater than 2) {
  var m = 0;
  while(m is less than or equal to a.length - n)
  {
    var first = m;
    var last = m + n - 1;
    var diff = a[last] - a[first];
    if (diff is equal to n - 1 && diff is greater than max) {
      Update max to be n;
      Break out of inner loop;
    }
    increment m;
    }
  decrement n;
}

console.log(max);

JSFiddle

Answer 7

Answer №4

One efficient approach could be to loop through the input array and compare each element with the previous one to find the longest range. Here is a sample implementation:

function findLongestRange(input) {
  let maxLength = 0
  let currentLength = 0

  for (let i = 0; i < input.length; i++) {
    if (i !== input.length) {
      if (input[i] === input[i + 1] - 1) {
        currentLength++
      } else {
        if (maxLength <= currentLength && currentLength !== 0) {
          maxLength = currentLength + 1
        }
        currentLength = 0
      }
    }
  }

  return maxLength
}

const data = [1, 2, 5, 9, 10, 12, 20, 21, 22, 23, 24, 26, 27]
console.log(findLongestRange(data))

To verify how this function performs with different inputs, here's a version of it with test cases included:

const data = [1, 2, 5, 9, 10, 12, 20, 21, 22, 23, 24, 26, 27]

function findLongestRange(input) {
  let maxLength = 0
  let currentLength = 0

  for (let i = 0; i < input.length; i++) {
    if (i !== input.length) {
      if (input[i] === input[i + 1] - 1) {
        currentLength++
      } else {
        if (maxLength <= currentLength && currentLength !== 0) {
          maxLength = currentLength + 1
        }
        currentLength = 0
      }
    }
  }

  return maxLength
}

console.clear()
;[
  [[1,2,5,6,7,1,2], 3],
  [[], 0],
  [data, 5],
  [[1,2,3], 3],
  [[1,3,4,6,8,1], 2],
  [[1,3,5], 0],
].forEach((test, index) => {
  const result = findLongestRange(test[0])
  console.assert(result === test[1], `Fail #${index}: Exp: ${test[1]}, got ${result}`)
})

Answer 8

One efficient approach could be to loop through the input array and compare each element with the previous one to find the longest range. Here is a sample implementation:

function findLongestRange(input) {
  let maxLength = 0
  let currentLength = 0

  for (let i = 0; i < input.length; i++) {
    if (i !== input.length) {
      if (input[i] === input[i + 1] - 1) {
        currentLength++
      } else {
        if (maxLength <= currentLength && currentLength !== 0) {
          maxLength = currentLength + 1
        }
        currentLength = 0
      }
    }
  }

  return maxLength
}

const data = [1, 2, 5, 9, 10, 12, 20, 21, 22, 23, 24, 26, 27]
console.log(findLongestRange(data))

To verify how this function performs with different inputs, here's a version of it with test cases included:

const data = [1, 2, 5, 9, 10, 12, 20, 21, 22, 23, 24, 26, 27]

function findLongestRange(input) {
  let maxLength = 0
  let currentLength = 0

  for (let i = 0; i < input.length; i++) {
    if (i !== input.length) {
      if (input[i] === input[i + 1] - 1) {
        currentLength++
      } else {
        if (maxLength <= currentLength && currentLength !== 0) {
          maxLength = currentLength + 1
        }
        currentLength = 0
      }
    }
  }

  return maxLength
}

console.clear()
;[
  [[1,2,5,6,7,1,2], 3],
  [[], 0],
  [data, 5],
  [[1,2,3], 3],
  [[1,3,4,6,8,1], 2],
  [[1,3,5], 0],
].forEach((test, index) => {
  const result = findLongestRange(test[0])
  console.assert(result === test[1], `Fail #${index}: Exp: ${test[1]}, got ${result}`)
})

Answer 9

Answer №5

An example of Python code:

l = [1,2,5,9,10,12,20,21,22,23,24,26,27]
current_range = None
current_range_val = 0
max_range = 0
max_range_val = 0
for i, j in zip(l, l[1:]):
    if j - i == 1:
        current_range_val += 1
        if current_range is None:
            current_range = (i, j)
        current_range = (current_range[0], j)
    else:
        if current_range_val > max_range_val:
            max_range = current_range
            max_range_val = current_range_val
        current_range_val = 0
        current_range = (j, None)

print(max_range)

This code snippet will output the range as:

(20, 24)

Answer 10

An example of Python code:

l = [1,2,5,9,10,12,20,21,22,23,24,26,27]
current_range = None
current_range_val = 0
max_range = 0
max_range_val = 0
for i, j in zip(l, l[1:]):
    if j - i == 1:
        current_range_val += 1
        if current_range is None:
            current_range = (i, j)
        current_range = (current_range[0], j)
    else:
        if current_range_val > max_range_val:
            max_range = current_range
            max_range_val = current_range_val
        current_range_val = 0
        current_range = (j, None)

print(max_range)

This code snippet will output the range as:

(20, 24)

Determining the largest range possible in a sorted array of integers

Answer №1

An Alternative Approach

Optimizing Performance

Answer №2

Answer №3

Answer №4

Answer №5

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