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README: Minimum Subarray Length to Remove for Divisibility

Problem Description

Given an array of integers nums and an integer p, the goal is to remove the smallest subarray from nums such that the sum of the remaining array is divisible by p. The task is to return the length of this subarray, or -1 if it's not possible.

Approach (Applicable to All Languages)

  1. Step 1: Calculate Total Sum
    Start by calculating the total sum of the array nums.

  2. Step 2: Find Remainder of Total Sum
    Find the remainder of the total sum when divided by p.

    • If the remainder is 0, the entire array is already divisible by p, and no subarray needs to be removed. Thus, return 0.

  3. Step 3: Track Prefix Sums Using a Hash Map
    Use a hash map to track prefix sums' remainders when divided by p. This helps in identifying subarrays that can be removed to achieve divisibility by p.

    • Initialize the hash map with {0: -1} to handle cases where the entire prefix contributes to divisibility.

  4. Step 4: Iterate Over the Array
    Loop through the array, maintaining a running sum (prefix sum) of the elements. For each element:

    • Compute the current prefix sum's remainder when divided by p.
    • Calculate the target remainder we need to achieve divisibility by p.

  5. Step 5: Check for Subarrays
    Check if the target remainder exists in the hash map:

    • If it exists, calculate the length of the subarray that can be removed, and update the minimum length of such subarrays found so far.
    • Update the hash map with the current prefix sum's remainder.

  6. Step 6: Return the Result
    After iterating through the array:

    • If a valid subarray was found, return the minimum length of such subarrays.
    • If no valid subarray was found, return -1.

C++ Code Explanation

  1. Initialize Variables: Start by calculating the total sum of the array and finding the remainder when divided by p.
  2. Handle Special Case: If the remainder is 0, return 0 as no subarray needs to be removed.
  3. Track Prefix Sums: Use an unordered map to track the remainder of prefix sums.
  4. Iterate Over the Array: For each element, calculate the running sum, find the corresponding target remainder, and check if it exists in the map.
  5. Update Result: If a valid subarray is found, update the minimum length.
  6. Return Result: Return the minimum length or -1 if no valid subarray is found.

Java Code Explanation

  1. Calculate Total Sum: Calculate the total sum of the elements in the array.
  2. Handle Special Case: If the total sum is already divisible by p, return 0.
  3. Use HashMap: Use a HashMap to track the remainders of prefix sums.
  4. Iterate Over Array: For each element, update the prefix sum and calculate the remainder. Check for the required target remainder in the map.
  5. Track Minimum Length: Keep track of the minimum length of the subarray that, if removed, makes the sum divisible by p.
  6. Return Final Result: Return -1 if no such subarray is found, otherwise return the minimum length.

JavaScript Code Explanation

  1. Calculate Total Sum: Sum up all the numbers in the array and find the remainder of the sum when divided by p.
  2. Handle Simple Case: If the remainder is 0, the entire array is divisible by p.
  3. Map for Prefix Modulo: Create a Map to store the remainders of the prefix sums.
  4. Iterate Through Array: For each element, update the prefix sum and calculate the remainder. Check for the target remainder in the map.
  5. Update Minimum Length: Whenever a valid subarray is found, calculate its length and update the minimum length.
  6. Return the Result: If no valid subarray is found, return -1.

Python Code Explanation

  1. Compute Total Sum: Calculate the total sum of the array and find the remainder when divided by p.
  2. Handle Divisibility Case: If the remainder is zero, return 0.
  3. Prefix Sum Tracking with Dictionary: Use a dictionary to track remainders of prefix sums and their indices.
  4. Iterate Over Elements: Compute the running sum and its remainder for each element. Check if the target remainder exists in the dictionary.
  5. Update Minimum Subarray Length: Whenever a matching remainder is found, update the minimum length of the subarray that needs to be removed.
  6. Return the Minimum Length or -1: After iterating, return the minimum subarray length or -1 if no valid subarray is found.

Go Code Explanation

  1. Total Sum Calculation: Start by calculating the total sum of the array and finding the remainder when divided by p.
  2. Handle Special Case: If the remainder is 0, return 0 since no subarray needs to be removed.
  3. Prefix Mod Tracking: Use a map to track prefix sums and their remainders.
  4. Iterate Through Array: For each element, update the prefix sum, calculate the current remainder, and check if the target remainder exists in the map.
  5. Update Minimum Length: If a valid subarray is found, update the minimum length.
  6. Return Result: Return the minimum length or -1 if no valid subarray is found.