Minimum Increment to Make Array Unique

class Solution {
public:
    int makeUnique(vector<int>& data) {
        int dataSize = data.size();
        int highestValue = 0;
        int requiredSteps = 0;

        // Compute the largest number in the vector data
        for (int number : data) {
            highestValue = max(highestValue, number);
        }

        // Construct a vector to count occurrences of each number in data
        vector<int> occurrenceCount(dataSize + highestValue, 0);

        // Fill occurrenceCount based on occurrence of numbers in data
        for (int number : data) {
            occurrenceCount[number]++;
        }

        // Adjust values to ensure all elements are unique
        for (int i = 0; i < occurrenceCount.size(); i++) {
            if (occurrenceCount[i] <= 1) continue;

            // Shift excess numbers to the next position
            int excess = occurrenceCount[i] - 1;
            occurrenceCount[i + 1] += excess;
            occurrenceCount[i] = 1;
            requiredSteps += excess;
        }

        return requiredSteps;
    }
};

This C++ solution implements a method to determine the minimum number of increments required to make all elements in an array unique. The approach involves several key steps:

1. Declare three variables: dataSize to store the size of the input vector data, highestValue to track the maximum value within data, and requiredSteps to keep count of the needed increments.

2. Loop through the data to find the highest value present. This will help in creating the occurrenceCount vector of appropriate size.

3. Create a vector occurrenceCount initialized with zeros, where its size is dataSize + highestValue + 1. This vector is used to count the occurrence of each value in the input vector.

4. Populate the occurrenceCount by incrementing the count at the index corresponding to every number in data.

5. Iterate through the occurrenceCount vector. For each position where the count of a number is more than one (indicating duplicates), calculate the excess by subtracting one from the count. Add the excess to the count at the next index (to shift the duplicates forward by increasing their value by one). Accumulate the value of excess into requiredSteps.

6. Return the accumulated requiredSteps, which represents the minimum increment required to make all array elements unique.

The advantage of this method is its efficiency in processing the uniqueness of the array elements through direct manipulation of count values rather than modifying the array itself. The use of the occurrenceCount ensures each element is processed in terms of its duplication status, handling situations effectively where multiple duplicates exist.

class Solution {

    public int minimumIncrementsToUnique(int[] elements) {
        int length = elements.length;
        int highest = 0;
        int totalIncrements = 0;

        // Finding the largest number in the array
        for (int num : elements) {
            highest = Math.max(highest, num);
        }

        // Array to count occurrences of each number
        int[] countArray = new int[length + highest];

        // Fill countArray with counts of each element
        for (int num : elements) {
            countArray[num]++;
        }

        // Process countArray to resolve duplicates
        for (int i = 0; i < countArray.length; i++) {
            if (countArray[i] <= 1) continue;

            // Reduce current count to 1 by moving excess to the next index
            int excess = countArray[i] - 1;
            countArray[i + 1] += excess;
            countArray[i] = 1;
            totalIncrements += excess;
        }

        return totalIncrements;
    }
}

The Java solution provided addresses the problem of making all elements in an array unique by incrementing duplicate elements. The approach utilizes an efficient method involving array manipulation to achieve this task. Below is a succinct summary of the solution's implementation strategy:

1. Calculate the length of the input array and identify the largest element to determine the size of an auxiliary count array.
2. Initialize the count array which keeps track of the occurrences of each number in the input array.
3. Traverse the count array:
   • Whenever a count exceeds one, implying a duplicate, calculate the excess.
   • Move the excess to the next index, effectively making each original number unique.
   • Maintain a cumulative total of increments made.
4. Return the total number of increments needed to make the array entries unique.

This method is space-efficient since it creates a count array large enough to accommodate all possible values of the array elements based on the maximum value found, and time-efficient as it resolves duplicates in a single pass through the count array.

class Solution:
    def uniqueIncrementsNeeded(self, array: List[int]) -> int:
        length = len(array)
        highest = max(array)
        extra_steps = 0

        # Array to track count of each number
        count_tracker = [0] * (length + highest)

        # Fill count array with occurrences of each number
        for number in array:
            count_tracker[number] += 1

        # Adjust counts to ensure all numbers are unique
        for index in range(len(count_tracker)):
            if count_tracker[index] <= 1:
                continue

            # Excess count needs to be moved to next index
            excess = count_tracker[index] - 1
            count_tracker[index + 1] += excess
            count_tracker[index] = 1
            extra_steps += excess

        return extra_steps

The Python solution provided addresses the problem of making all elements in an array unique through minimum increments. The code implements a function uniqueIncrementsNeeded within the Solution class that accepts an integer array and returns the minimum number of increments required to ensure all elements are unique. Here's how the solution works:

• Determine the length of the input array and locate its maximum value for creating a sufficient range in the count tracker.
• Initialize a count_tracker array that will record the occurrence of each element in the given array. The size of this count array is determined by the length of the input array added to its maximum value.
• Populate the count_tracker by iterating through the input array and incrementing the count at the indices matching the array values.
• Traverse the count_tracker and whenever a count exceeds one (indicating duplicates), reduce the count to one and transfer the excess count to the next index. This step also accumulates the number of steps taken to remove duplicates.
• Return the total count of required increments stored in extra_steps.

This solution efficiently ensures that all numbers become unique by incrementally adjusting duplicates and is effective for arrays where the elements and their count vary widely.