onenoughtone

Solution Approach

Solution Overview

There are 3 main approaches to solve this problem:

Dynamic Programming Approach - Complex approach
Optimized Dynamic Programming with Binary Search - Complex approach
State Compression with TreeMap - Complex approach

Approach 1: Dynamic Programming Approach

The key insight for solving this problem is to use dynamic programming to track the minimum number of replacements needed to make the array strictly increasing up to each position.

Let's define dp[i][j] as the minimum number of replacements needed to make the first i elements of arr1 strictly increasing, where the last element is j. Here, j can be either the original element from arr1 or a replacement from arr2.

For each position i, we have two options:

Keep the original element arr1[i-1].
Replace it with an element from arr2.

We need to ensure that the element at position i is greater than the element at position i-1.

Algorithm:

Sort arr2 and remove duplicates to simplify the problem.
Initialize a 2D DP table where dp[i][j] represents the minimum number of replacements needed to make the first i elements of arr1 strictly increasing, with the last element being j.
Initialize dp[0][j] = 0 for all j, as an empty array is strictly increasing.
For each position i from 1 to n:
a. For each possible value j at position i-1:
i. If arr1[i-1] > j, update dp[i][arr1[i-1]] = min(dp[i][arr1[i-1]], dp[i-1][j]) (keep the original element).
ii. For each element k in arr2 that is greater than j, update dp[i][k] = min(dp[i][k], dp[i-1][j] + 1) (replace with an element from arr2).
Return the minimum value in the last row of the DP table, or -1 if it's impossible to make arr1 strictly increasing.

Time Complexity:

O(n * m²)

Where n is the length of arr1 and m is the length of arr2. For each position in arr1, we consider all possible values from arr2 as the previous and current elements.

Space Complexity:

O(n * m)

We need a 2D DP table of size n×m to store the minimum number of replacements for each position and value.

Approach 2: Optimized Dynamic Programming with Binary Search

We can optimize the previous approach by using binary search to find the smallest element in arr2 that is greater than a given value. This reduces the time complexity from O(n * m²) to O(n * m * log m).

Additionally, we can use a 1D DP array instead of a 2D array by keeping track of the minimum number of replacements needed for each possible last value.

Algorithm:

Sort arr2 and remove duplicates.
Initialize a map dp where dp[j] represents the minimum number of replacements needed to make arr1 strictly increasing with the last element being j.
Initialize dp[arr1[0]] = 0 (keep the original first element) and dp[arr2[i]] = 1 for all arr2[i] (replace the first element).
For each position i from 1 to n-1:
a. Initialize a new map newDp to store the updated DP values.
b. For each entry (prev, count) in dp:
i. If arr1[i] > prev, update newDp[arr1[i]] = min(newDp[arr1[i]], count) (keep the original element).
ii. Use binary search to find the smallest element in arr2 that is greater than prev, and update newDp[arr2[j]] = min(newDp[arr2[j]], count + 1) (replace with an element from arr2).
c. If newDp is empty, return -1 (it's impossible to make arr1 strictly increasing).
d. Update dp = newDp.
Return the minimum value in dp, or -1 if dp is empty.

Time Complexity:

O(n * m * log m)

Where n is the length of arr1 and m is the length of arr2. For each position in arr1, we consider all possible values from the previous position (at most m) and use binary search (log m) to find the smallest valid replacement.

Space Complexity:

O(m)

We only need to store the minimum number of replacements for each possible last value, which is at most m.

Approach 3: State Compression with TreeMap

We can further optimize the space complexity by using a TreeMap (or a similar data structure) to store only the necessary states. This approach is particularly useful when the range of possible values is large.

The key idea is to use the value as the key in the TreeMap and the minimum number of replacements as the value. This allows us to efficiently find the smallest value greater than a given value (using the ceiling function) and update the minimum number of replacements.

Algorithm:

Sort arr2 and remove duplicates.
Initialize a TreeMap dp where dp[j] represents the minimum number of replacements needed to make arr1 strictly increasing with the last element being j.
Initialize dp[arr1[0]] = 0 (keep the original first element) and dp[arr2[i]] = 1 for all arr2[i] < arr1[0] (replace the first element).
For each position i from 1 to n-1:
a. Initialize a new TreeMap newDp to store the updated DP values.
b. For each entry (prev, count) in dp:
i. If arr1[i] > prev, update newDp[arr1[i]] = min(newDp[arr1[i]], count) (keep the original element).
ii. Find the smallest element in arr2 that is greater than prev using the ceiling function, and update newDp[arr2[j]] = min(newDp[arr2[j]], count + 1) (replace with an element from arr2).
c. If newDp is empty, return -1 (it's impossible to make arr1 strictly increasing).
d. Update dp = newDp.
Return the minimum value in dp, or -1 if dp is empty.

Time Complexity:

O(n * m * log m)

Where n is the length of arr1 and m is the length of arr2. The time complexity is the same as the previous approach, but the constant factors are better due to the use of a TreeMap.

Space Complexity:

O(m)

We only need to store the minimum number of replacements for each possible last value, which is at most m.

Pseudocode

solution.pseudo

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function makeArrayIncreasing(arr1, arr2):
    // Sort arr2 and remove duplicates
    arr2 = sort(arr2)
    arr2 = removeDuplicates(arr2)
 
    n = length of arr1
    m = length of arr2
 
    // Initialize DP table with infinity
    dp = 2D array of size (n+1) × (10^9+1), initialized with infinity
 
    // Base case: empty array is strictly increasing
    dp[0][0] = 0
 
    for i from 1 to n:
        for j from 0 to 10^9:
            // Option 1: Keep the original element
            if arr1[i-1] > j:
                dp[i][arr1[i-1]] = min(dp[i][arr1[i-1]], dp[i-1][j])
 
            // Option 2: Replace with an element from arr2
            for k in arr2:
                if k > j:
                    dp[i][k] = min(dp[i][k], dp[i-1][j] + 1)
 
    // Find the minimum number of replacements
    minReplacements = infinity
 
    for j from 0 to 10^9:
        minReplacements = min(minReplacements, dp[n][j])
 
    return minReplacements == infinity ? -1 : minReplacements

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Learning Objective

Understand different approaches to solve the array increaser problem and analyze their efficiency.