Smallest Missing Non-negative Integer After Operations - Complete Solution Guide

Smallest Missing Non-negative Integer After Operations is LeetCode problem 2598, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3.

Problem Statement

You are given a 0-indexed integer array nums and an integer value . In one operation, you can add or subtract value from any element of nums . For example, if nums = [1,2,3] and value = 2 , you can choose to subtract value from nums[0] to make nums = [-1,2,3] . The MEX (minimum excluded) of an array is the smallest missing non-negative integer in it. For example, the MEX of [-1,2,3] is 0 while the MEX of [1,0,3] is 2 . Return the maximum MEX of nums after applying the mentioned operation any num

Detailed Explanation

The problem asks us to find the smallest non-negative integer (MEX) that is missing from an array `nums` after we can apply an operation any number of times to each element. The operation involves adding or subtracting a given integer `value` from any element. Crucially, we're trying to *maximize* the MEX achievable after these operations. The input is an array of integers `nums` and an integer `value`. The output is the maximum achievable MEX.

Solution Approach

The provided solution uses the insight that the remainder modulo `value` is invariant under the allowed operations. It counts the number of elements in `nums` that have each possible remainder modulo `value`. Then, it iterates from 0, checking if each number `mex` can be 'constructed' using the elements of `nums`. If the count for the remainder `mex % value` is greater than 0, it means we can use an element from `nums` to 'make' the number `mex`. If the count is 0, then `mex` is the smallest missing integer, and we return it.

Step-by-Step Algorithm

1. Step 1: Initialize a list `counts` of size `value` with all elements set to 0. This list will store the counts of each remainder modulo `value`.
2. Step 2: Iterate through the `nums` array. For each number `num`, calculate `num % value`. Since `num` can be negative, we can use `(num % value + value) % value` to ensure a non-negative remainder. Increment the count at the corresponding index in the `counts` list.
3. Step 3: Initialize `mex` to 0. This variable represents the smallest missing non-negative integer.
4. Step 4: Enter a `while` loop that continues until the MEX is found.
5. Step 5: Inside the loop, calculate the remainder `remainder_needed` as `mex % value`.
6. Step 6: Check if the count at `counts[remainder_needed]` is greater than 0. If it is, it means we can 'make' the number `mex` by applying the allowed operations. Decrement the count `counts[remainder_needed]` and increment `mex` to check the next number.
7. Step 7: If the count at `counts[remainder_needed]` is 0, it means we cannot 'make' the number `mex`, so `mex` is the smallest missing non-negative integer. Return `mex`.

Key Insights

• Insight 1: The operation of adding or subtracting `value` doesn't change the element's remainder when divided by `value`. This is because (a + k*value) % value == a % value.
• Insight 2: We can group elements of `nums` based on their remainders modulo `value`. Each remainder group can contribute at most one of each non-negative integer with that same remainder.
• Insight 3: The MEX can be determined greedily by checking for the smallest missing number starting from 0, 1, 2,... and using the counts of remainders to 'construct' these numbers using allowed operations.

Complexity Analysis

Time Complexity: O(N + value), where N is the length of the `nums` array and value is the given integer. The loop iterates at most `value` times to find the MEX. Building counts array takes O(N) time.

Space Complexity: O(value), as we use a list `counts` of size `value` to store the counts of remainders.

Topics

This problem involves: Array, Hash Table, Math, Greedy.

Companies

Asked at: Atlassian, IBM, Mercari.