Count Number of Bad Pairs

class Solution {
public:
    long long countIncorrectPairs(vector<int>& elements) {
        long long badPairCount = 0;
        unordered_map<int, int> diffMap;

        for (int index = 0; index < elements.size(); index++) {
            int difference = index - elements[index];
            
            int countOfGoodPairs = diffMap[difference];
            
            badPairCount += index - countOfGoodPairs;
            
            diffMap[difference] = countOfGoodPairs + 1;
        }

        return badPairCount;
    }
};

The given C++ program aims to calculate the number of "bad pairs" in a vector of integers labelled as elements. A bad pair is defined as a pair of indices (i, j) where i < j and i - elements[i] does not equal j - elements[j].

• The method countIncorrectPairs is defined to compute the number of bad pairs.
• Inside this method, a variable badPairCount is used to maintain the count of bad pairs found during the execution.
• An unordered_map named diffMap stores the difference index - elements[index] for each index, alongside the count of how many times this difference has occurred.
• The program efficiently manages the computation by iterating through each element of the array:
  • For each index, it calculates the difference, index - elements[index].
  • Using this difference, the program checks how many times this difference has previously appeared (countOfGoodPairs from diffMap).
  • The number of bad pairs is then incremented by the total number of previous indices minus the number of times the current computed difference has occurred. This calculation ensures that the reverse scenarios are accounted for.
  • The difference count for this index is updated in diffMap.

This method efficiently calculates the count of bad pairs by leveraging the difference between the indices and their respective values, ensuring that the need for comparing every possible pair (which is computationally expensive) is avoided. This results in a more optimized solution in terms of both time and space complexities.

class Solution {

    public long calculateBadPairs(int[] values) {
        long badPairCount = 0;
        Map<Integer, Integer> diffMap = new HashMap<>();

        for (int index = 0; index < values.length; index++) {
            int difference = index - values[index];
            int existCount = diffMap.getOrDefault(difference, 0);

            badPairCount += index - existCount;
            diffMap.put(difference, existCount + 1);
        }

        return badPairCount;
    }
}

The Java solution provided calculates the number of bad pairs in an array of integers. A pair (i, j) is defined as bad if (i < j) yet j - values[j] != i - values[i]. The problem is approached by using a HashMap to keep track of frequency of the differences between index positions and their corresponding values.

Here is a breakdown of how the solution is implemented:

• Initialize badPairCount to keep a running total of bad pairs.
• Create a diffMap HashMap to store frequencies of each difference between indices and values.
• Loop through the array using an index. For each element:
  • Calculate difference as (index - values[index]).
  • Retrieve the count of this difference from diffMap using getOrDefault(), which fetches the count if it exists, or returns 0 if it doesn't.
  • Increment the badPairCount by (index - existCount). This represents the number of new bad pairs found at the current index.
  • Update diffMap by increasing the count of the current difference.
• Return badPairCount.

This approach ensures optimal performance by avoiding the brute force method of comparing each pair, leveraging the properties of differences to find and count bad pairs efficiently.

class Solution:
    def countNonGoodPairs(self, elements: List[int]) -> int:
        non_good_pairs = 0
        diff_tracker = {}

        for index in range(len(elements)):
            current_diff = index - elements[index]

            # Existing counts of good pairs with same difference
            current_good_pairs = diff_tracker.get(current_diff, 0)

            # Calculate non-good pairs by subtracting number of good pairs from possible pairs
            non_good_pairs += index - current_good_pairs

            # Update our tracker for this current difference
            diff_tracker[current_diff] = current_good_pairs + 1

        return non_good_pairs

The Python3 code provided effectively counts the number of non-good pairs in an array where a pair (i, j) is considered bad if i < j and i - j ≠ elements[i] - elements[j]. It uses a dictionary to track the frequency of differences (index - elements[index]), which helps identify the good pairs.

Here's a breakdown of the solution logic:

• Initialize a variable non_good_pairs to zero to store the count of non-good pairs.
• Use a dictionary diff_tracker to keep track of the differences and their counts.
• Iterate through the array elements using the index.
• Compute the difference between the current index and the value at that index.
• Fetch the current count of good pairs that have the same calculated difference from diff_tracker.
• Calculate non-good pairs by subtracting the good pairs from the total possible pairs up to the current index.
• Update the count of the current difference in the tracker.
• Finally, the function returns the total non_good_pairs.

This approach efficiently utilizes a dictionary to avoid the direct computation of each pair, thus optimizing the solution by minimizing redundant operations and making use of past computations.