Problem Statement
The goal is to analyze an integer array arr and determine how many elements x in the array have the characteristic that the element x + 1 is also present in the array. It's crucial to note that if there are duplicates of x in arr, each duplicate should be counted separately. This implies that every instance of an element that satisfies the condition should influence the final count independently of its duplicates.
Examples
Example 1
Input:
arr = [1,2,3]
Output:
2
Explanation:
1 and 2 are counted cause 2 and 3 are in arr.
Example 2
Input:
arr = [1,1,3,3,5,5,7,7]
Output:
0
Explanation:
No numbers are counted, cause there is no 2, 4, 6, or 8 in arr.
Constraints
1 <= arr.length <= 10000 <= arr[i] <= 1000
Approach and Intuition
The problem is straightforward where the task is to find the count of such elements x in the array arr that also have their consecutive number x + 1 present in the array. Here's how you can approach solving this problem:
- Read the input array
arr. - Create a structure to keep track of all elements in the array for quick lookup. The best fit would be a set since it allows O(1) average-time complexity for lookups and avoids counting the same element more than once.
- Initialize a counter to zero. This will hold the count of elements meeting the condition.
- Loop through each element
xin the array:- For every element
x, check ifx + 1exists in the set. - If
x + 1is found, increment the counter.
- For every element
- After iterating through the array, the counter will represent the total number of elements
xsuch that bothxandx + 1are present.
Examples Revisited
To offer further intuition, let's revisit the provided examples considering the above approach:
For Example 1, the array
arris [1,2,3]. Both 1 and 2 have their consecutive numbers 2 and 3, respectively, present in the array. Hence, the counter results in 2.In Example 2, the array is [1,1,3,3,5,5,7,7]. The consecutive elements after each number in the array, such as 2, 4, 6, and 8, are not present in the array. As a result, the counter remains 0.
This procedure ensures effective counting while maintaining a low time complexity, leveraging the properties of set data structures for fast element lookup.
Solutions
- Java
- Python
public int tallyRepetitiveElements(int[] data) {
Arrays.sort(data);
int tally = 0;
int sequenceLength = 1;
for (int index = 1; index < data.length; index++) {
if (data[index - 1] != data[index]) {
if (data[index - 1] + 1 == data[index]) {
tally += sequenceLength;
}
sequenceLength = 0;
}
sequenceLength++;
}
return tally;
}
The given Java code defines a method tallyRepetitiveElements that computes the tally of elements followed directly by the next consecutive integer in a sorted array. Here's a breakdown of the code:
- Begin by sorting the input array using
Arrays.sort(data). - Initialize two variables:
tallyto track the count of elements meeting the criteria, andsequenceLengthto handle the length of consecutive runs. - Iterate through the array starting from the second element to compare with its predecessor.
- Check conditions:
- If the current element is different from the previous one (
data[index - 1] != data[index]), then check if the previous element and current element are consecutive numbers (data[index - 1] + 1 == data[index]). If true, add the sequence length to the tally. - Reset
sequenceLengthif elements are not the same. - Increment
sequenceLengthafter each iteration for assessing the next potential sequence.
- If the current element is different from the previous one (
- Return the tally which includes the count of elements immediately followed by their consecutive integer in the array.
By following this approach, you effectively analyze and count sequences of consecutive numbers in an array.
def tallyElements(self, nums: List[int]) -> int:
nums.sort()
tally = 0
sequence_len = 1
for index in range(len(nums)):
if nums[index - 1] != nums[index]:
if nums[index - 1] + 1 == nums[index]:
tally += sequence_len
sequence_len = 0
sequence_len += 1
return tally
This Python function tallyElements calculates the count of elements that have a next consecutive number in a given list nums. The approach involves sorting the list initially, which helps in linearly checking consecutive elements. Here's how the function operates:
- Firstly, sort the
numslist. This ensures that all consecutive numbers are adjacent, facilitating easier counting. - Initialize
tallyto store the number of valid elements andsequence_lento count consecutive duplicates. - Iterate through each element in the list
numsstarting from the first index. Use a for loop that iterates based on the index of list elements. - Inside the loop, check if the current element and the previous element are consecutive (i.e., the difference between them is exactly 1). If they are, increment the
tallybysequence_lenwhich counts the current sequence of consecutive duplicates. - Reset
sequence_lento zero when the current and previous elements are not the same, ensuring that the next sequence count starts anew. - Finally, return
tallyas the output of the function.
The returned integer represents the number of elements in the list which are immediately followed by their consecutive number. This function effectively leverages sorting and linear traversal to solve the problem efficiently.
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