2121. Intervals Between Identical Elements

Problem Description

The problem provides us with an array arr where each element may occur more than once. For each element arr[i], we want to calculate the sum of distances (intervals) between arr[i] and all other elements in arr that have the same value as arr[i]. In other words, for every element that matches arr[i], we sum up the absolute differences in their indices. The output should be an array intervals of the same length as arr where each intervals[i] contains the computed sum for arr[i].

To put it simply, if an element appears multiple times in the array, we are looking to find out how far apart all of those occurrences are from each other, summing up these distances for each unique value.

Intuition

The solution approach makes use of a dictionary (or equivalent) to store occurrences of each unique value in the given array. Keys in this dictionary are the unique values, and the values are lists containing the indices where the key occurs.

The overarching idea is to use these lists to calculate the intervals in an efficient manner. A neat observation is that when calculating the sum of intervals for a particular number in arr, we notice that if the indices are sorted, each index v[i] will contribute (i * v[i]) - ((m - i) * v[i]) to the total sum, where i is the current position within the occurrences of the number in arr, and m is the total number of occurrences. Put differently, for any given occurrence, the distances from it to all previous occurrences will each increase by v[i] - v[i - 1] when moving to the next occurrence, and the distances to all next occurrences will each decrease by the same amount.

The algorithm's efficiency comes from leveraging this pattern to update the sum iteratively while traversing through the list of occurrences of each unique value, thereby avoiding the need to recompute the sum from scratch for every element.

Here is a step-by-step breakdown of the solution:

1. Create a dictionary (d) to group elements of arr by their values, mapping each value to a list of indices at which it occurs in arr.
2. Initialize an answer array (ans) of size n with zeros to hold the sum of intervals for each corresponding element in arr.
3. Iterate through the grouped values in the dictionary:
   • For the current group, calculate the initial interval sum for the first occurrence.
   • Iterate through the indices in the current group:
     • Calculate the difference between the current index and the previous one.
     • Update the running total (val), adding the interval increments for indices before the current one and subtracting for indices after.
     • Store the running total in the corresponding position in the ans array.
4. Return the ans array, which now contains the sum of intervals for each element in arr.

By organizing the elements in this way and cleverly calculating the contributions of each element's occurrences to the sum, the solution achieves efficient computation of the intervals.

Learn more about Prefix Sum patterns.

Solution Approach

The solution approach can be broken down into the following steps:

1. Group Elements by Value: The defaultdict(list) is used to create a dictionary d where each key is a value in the array arr, and the corresponding value is a list of indices at which this key appears. The enumerate function is used to get both the index i and the value v as we iterate through arr. This step essentially groups the occurrences of each element for easy access.

   1d = defaultdict(list)
   2for i, v in enumerate(arr):
   3    d[v].append(i)

2. Prepare the Answer Array: An array ans of size n (where n is the length of the input array arr) is initialized with zeros. This will be used to store the computed sum of intervals for each element within arr.

   1ans = [0] * n

3. Iterate Over the Groups: Then, we iterate through all lists of indices (all values) present in d. For each list v, we calculate the sum of intervals for the first element in the list as val. Here, m refers to the total number of occurrences in the list v.

   1for v in d.values():
   2    m = len(v)
   3    val = sum(v) - v[0] * m

4. Compute the Intervals Using a Running Total: As we iterate over the indices in v:

   • We incrementally update val by adding the cumulative distance going forward and subtracting the cumulative distance going backwards in the list of occurrences.
   • The delta represents the difference between the current index and the previous one, which affects the cumulative distances in this manner.
   • The final sum for each occurrence p of the value is then saved in ans[p].

   1for i, p in enumerate(v):
   2    delta = v[i] - v[i - 1] if i >= 1 else 0
   3    val += i * delta - (m - i) * delta
   4    ans[p] = val

5. Return the Result: The ans array is returned, which holds the sum of intervals for each element of the original array arr.

The solution approach combines dictionary mapping with efficient handling of sequential updates, thereby avoiding redundant calculations. This reduces the complexity considerably compared to a naive approach that might involve nested loops. The use of a running total (incremental adjustment of the val) to calculate the distances as we iterate through each list of occurrences is key to the efficiency of the solution.

Example Walkthrough

Let's consider a small example to illustrate the solution approach. Suppose we have the following array arr:

1arr = [1, 3, 1, 3, 2, 1]

First, we want to group the elements by value. We will get a dictionary d that looks like this:

1d = {
2  1: [0, 2, 5],
3  3: [1, 3],
4  2: [4]
5}

Here, the keys of the dictionary are the unique values from arr, and the values are lists containing the indices at which the corresponding key occurs in arr.

The answer array ans will initially be all zeros:

1ans = [0, 0, 0, 0, 0, 0]  # since there are 6 elements in arr

We iterate over the entries in d:

• For the group with key 1, v is [0, 2, 5]. The length of v (m) is 3. The initial interval sum val is calculated as val = sum([0, 2, 5]) - (0 * 3) = 7.

• Within the group:

  • The difference (delta) between the first and second occurrence of 1 is 2 - 0 = 2. Update val to val + (1 * 2) - (2 * 2) = 7 + 2 - 4 = 5. ans[2] is set to 5.
  • The difference between the second and third occurrence is 5 - 2 = 3. Update val to val + (2 * 3) - (1 * 3) = 5 + 6 - 3 = 8. ans[5] is set to 8.

• Moving on to the group with key 3, v is [1, 3]. Here, m is 2 and initial val is val = sum([1, 3]) - (1 * 2) = 4 - 2 = 2.

• Within the group:

  • The difference (delta) between the first and second occurrence of 3 is 3 - 1 = 2. Update val to val + (1 * 2) - (1 * 2) = 2 + 2 - 2 = 2. ans[1] and ans[3] are both set to 2 since this is the total distance for both occurrences.

• Finally, for the group with key 2, v is [4]. Because there is only one occurrence, the distance is zero. ans[4] remains 0.

The final ans array will be:

1ans = [0, 2, 5, 2, 0, 8]

In this final array, each ans[i] is the sum of intervals between the occurrences of arr[i] in the original array. This efficiently solves the problem by using a dictionary to map values to indices and by calculating the sum incrementally.

The solution leverages the spatial ordering of index occurrences, making use of a running total instead of performing redundant calculations, which would happen if doing this for each element individually without considering the previously calculated distances.

Solution Implementation

Time and Space Complexity

The given Python code involves creating a dictionary to group indices of identical elements and then computing the sum of distances for each unique element in the array.

Time Complexity:

1. Constructing the dictionary d where each value is a list of indices has a time complexity of O(n), where n is the length of the array arr. This is because we iterate over the array once.

2. The outer loop for v in d.values(): has at most n iterations in the worst-case scenario where all array elements are unique.

3. Inside this loop, we calculate the initial val for each group of identical elements. The summation sum(v) is O(m), where m is the number of times a distinct element appears in the array. The term v[0] * m is O(m) since it involves m multiplications. As such, this part has a time complexity of O(m) for each unique element.

4. The inner loop for i, p in enumerate(v): iterates m times (where m is the size of the list v). The calculation within this loop is O(1), as it consists of simple arithmetic operations. Therefore, the time complexity for the inner loop is O(m).

Combining these observations, the overall time complexity of the algorithm depends on the distribution of elements in the array. In the worst-case scenario, where all elements are unique, the time complexity tends to O(n * m) which simplifies to O(n). In the average and expected case, where elements are repeated, the time complexity would be O(n + k * m) where k is the number of unique elements. However, since the sum of all m for each unique element is n, the average-case time complexity remains linear, or O(n).

Space Complexity:

1. The space complexity of the dictionary d can be up to O(n) in the case where all elements are distinct since we store an index for each element.

2. The ans list also takes up O(n) space since it stores the result for each element in the arr.

3. Auxiliary space complexity is O(1) as there are no other data structures that grow with input size; only a fixed number of variables are used.

Therefore, the space complexity of the code is O(n) accounting for the space required for the dictionary d and the answer array ans.

Learn more about how to find time and space complexity quickly using problem constraints.