945. Minimum Increment to Make Array Unique

[mah-shamim]

You are given an integer array nums. In one move, you can pick an index i where 0 <= i < nums.length and increment nums[i] by 1.

Return the minimum number of moves to make every value in nums unique.

The test cases are generated so that the answer fits in a 32-bit integer.

Example 1:

• Input: nums = [1,2,2]
• Output: 1
• Explanation: After 1 move, the array could be [1, 2, 3].

Example 2:

• Input: nums = [3,2,1,2,1,7]
• Output: 6
• Explanation: After 6 moves, the array could be [3, 4, 1, 2, 5, 7].
  It can be shown with 5 or less moves that it is impossible for the array to have all unique values.

Constraints:

• 1 <= nums.length <= 105
• 0 <= nums[i] <= 105

[mah-shamim]

The problem asks us to make every number in a given integer array unique. We can increment any number in the array in one move, and the goal is to minimize the total number of moves required to ensure that each element is distinct.

Key Points:

• You are allowed to increment numbers in the array.
• The task is to find the minimum number of moves necessary to make all numbers unique.
• You need to handle potentially large arrays, so efficiency is important.

Approach:

The strategy involves sorting the array and then iterating over it, ensuring each number is greater than the last (if a number repeats). If a number is smaller than or equal to the previous number (or the minimum available number), we increment it and count the moves.

• Sorting the array ensures that we process the numbers in ascending order, making it easy to see if a number is already taken by a previous one.
• We maintain a variable minAvailable which tracks the smallest number that can be assigned to a number (starting from 0).
• If a number in the array is less than minAvailable, we must increment it to minAvailable and update the counter for moves.

Plan:

1. Sort the array to arrange the numbers in increasing order.
2. Initialize two variables:

• ans: To store the total number of moves.
• minAvailable: To track the next available unique number.

3. Loop through the sorted array:

• For each number, if it's less than minAvailable, increment it until it's unique.
• Update minAvailable to be the next available number.

4. Return the total number of moves.

Let's implement this solution in PHP: 945. Minimum Increment to Make Array Unique

<?php
/**
 * @param Integer[] $nums
 * @return Integer
 */
function minIncrementForUnique($nums) {
    $ans = 0;
    $minAvailable = 0;

    sort($nums);

    foreach ($nums as $num) {
        $ans += max($minAvailable - $num, 0);
        $minAvailable = max($minAvailable, $num) + 1;
    }

    return $ans;
}

// Example usage:
$nums1 = [1,2,2];
$nums2 = [3,2,1,2,1,7];

echo minIncrementForUnique($nums1) . "\n"; // Output: 1
echo minIncrementForUnique($nums2) . "\n"; // Output: 6
?>

Explanation:

1. Sorting: The array is sorted so that we can easily compare each number to the previous one.
2. Loop: For each number, if it's smaller than minAvailable, it needs to be incremented. We update minAvailable to be one more than the current number after it's incremented.
3. Counting moves: Every time a number is incremented to make it unique, we add to the total moves.

Example Walkthrough:

Example 1:

• Input: nums = [1, 2, 2]
• Sorted: [1, 2, 2]
• Initial minAvailable: 0
  • First number (1): No increment needed. minAvailable becomes 2.
  • Second number (2): No increment needed. minAvailable becomes 3.
  • Third number (2): Since minAvailable is 3, we increment 2 to 3. So, the total moves = 1.
• Output: 1

Example 2:

• Input: nums = [3, 2, 1, 2, 1, 7]
• Sorted: [1, 1, 2, 2, 3, 7]
• Initial minAvailable: 0
  • First number (1): No increment needed. minAvailable becomes 2.
  • Second number (1): Increment it to 2. Total moves = 1. minAvailable becomes 3.
  • Third number (2): Increment it to 3. Total moves = 2. minAvailable becomes 4.
  • Fourth number (2): Increment it to 4. Total moves = 4. minAvailable becomes 5.
  • Fifth number (3): Increment it to 5. Total moves = 6. minAvailable becomes 6.
  • Sixth number (7): No increment needed. minAvailable becomes 8.
• Output: 6

Time Complexity:

• Sorting: The sorting operation takes O(n log n), where n is the length of the array.
• Iteration: The iteration through the array is O(n).
  Thus, the overall time complexity is O(n log n), dominated by the sorting step.

Output for Example:

Example 1:

• Input: [1, 2, 2]
• Output: 1

Example 2:

• Input: [3, 2, 1, 2, 1, 7]
• Output: 6

The approach efficiently solves the problem by sorting the array and using a greedy strategy to make each element unique. By processing the elements in ascending order, we ensure that we minimize the number of moves required to resolve duplicates. The solution has a time complexity of O(n log n), making it feasible even for large inputs (up to 10⁵ elements).

[basharul-siddike]

Thanks to solve this problem

[mah-shamim]

Thanks