Problem Description

You are given two integer arrays arr1 and arr2. Your goal is to make arr1 strictly increasing using the minimum number of operations.

In each operation, you can:

• Choose any index i from arr1 (where 0 <= i < arr1.length)
• Choose any index j from arr2 (where 0 <= j < arr2.length)
• Replace arr1[i] with arr2[j]

An array is strictly increasing if each element is strictly greater than the previous one (i.e., arr[i] < arr[i+1] for all valid i).

The function should return the minimum number of operations needed to make arr1 strictly increasing. If it's impossible to achieve this, return -1.

For example:

• If arr1 = [1, 5, 3, 6, 7] and arr2 = [1, 3, 2, 4], you could replace arr1[2] (which is 3) with arr2[3] (which is 4) to get [1, 5, 4, 6, 7]. But this still isn't strictly increasing because 5 > 4. You'd need to make additional replacements.
• The key challenge is finding the optimal sequence of replacements that minimizes the total number of operations while ensuring the final array is strictly increasing.

Intuition

The key insight is that we need to decide for each position whether to keep the original value or replace it with a value from arr2. This is a classic dynamic programming problem where we need to track the minimum operations needed to make the array strictly increasing up to each position.

First, let's think about what makes this problem tractable. If we're at position i and we want to keep arr1[i], we need the previous elements to be strictly less than arr1[i]. This means we might need to replace some previous elements to ensure this property holds.

The critical observation is that when we replace elements, we should replace them with the largest possible values that still maintain the strictly increasing property. Why? Because using larger values gives us more flexibility for future positions - it's easier to find values greater than what we've already placed.

This leads us to consider: for each position i, what if we keep arr1[i]? We need to look back and see how many consecutive elements before i we might need to replace. If we replace k consecutive elements before position i, we need to:

1. Find k values from arr2 that form a strictly increasing sequence
2. These values must be less than arr1[i]
3. The element at position i-k-1 (if it exists) must be less than the first replacement value

To efficiently find replacement values, we sort arr2 and remove duplicates. When looking for replacements, we use binary search to find the largest possible values from arr2 that are still less than arr1[i].

The dynamic programming state f[i] represents the minimum operations needed to make arr1[0...i] strictly increasing while keeping arr1[i] unchanged. By adding sentinels -inf and inf at the beginning and end, we ensure that the first and last elements are never replaced, simplifying our logic. The answer will be f[n-1] since the last sentinel inf ensures any valid sequence before it will be strictly increasing.

Learn more about Binary Search, Dynamic Programming and Sorting patterns.

Solution Approach

The implementation follows these key steps:

1. Preprocessing arr2:

arr2.sort()
m = 0
for x in arr2:
    if m == 0 or x != arr2[m - 1]:
        arr2[m] = x
        m += 1
arr2 = arr2[:m]

We sort arr2 and remove duplicates to create a pool of unique replacement values in ascending order. This allows us to use binary search efficiently later.

2. Adding Sentinels:

arr = [-inf] + arr1 + [inf]
n = len(arr)

We add -inf at the beginning and inf at the end as sentinels. This ensures:

• The first element is never replaced (it's already the smallest possible)
• The last element is never replaced (it's already the largest possible)
• We can focus on making the array strictly increasing up to the last real element

3. Dynamic Programming Setup:

f = [inf] * n
f[0] = 0

Initialize f[i] to represent the minimum operations needed to make arr[0...i] strictly increasing while keeping arr[i] unchanged. Set f[0] = 0 since the sentinel -inf requires no operations.

4. Main DP Transition: For each position i from 1 to n-1:

if arr[i - 1] < arr[i]:
    f[i] = f[i - 1]

First, check if we can keep both arr[i-1] and arr[i] without any replacements between them. If they already form an increasing pair, we can transfer the state directly.

j = bisect_left(arr2, arr[i])
for k in range(1, min(i - 1, j) + 1):
    if arr[i - k - 1] < arr2[j - k]:
        f[i] = min(f[i], f[i - k - 1] + k)

Then, consider replacing k consecutive elements before position i:

• Use binary search to find index j - the first position in arr2 where the value is >= arr[i]
• Try replacing k elements (from 1 to min(i-1, j)) before position i
• For each k, we need k strictly increasing values from arr2 that are all less than arr[i]
• These values would be arr2[j-k], arr2[j-k+1], ..., arr2[j-1] (the largest k values less than arr[i])
• Check if arr[i-k-1] < arr2[j-k] to ensure the sequence remains strictly increasing
• If valid, update f[i] with the minimum of current value and f[i-k-1] + k

5. Return Result:

return -1 if f[n - 1] >= inf else f[n - 1]

If f[n-1] is still inf, it means there's no way to make the array strictly increasing. Otherwise, return the minimum number of operations.

The time complexity is O(n^2 + m log m) where n is the length of arr1 and m is the length of arr2. The space complexity is O(n + m).

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Example: arr1 = [1, 5, 3, 6] and arr2 = [4, 3, 1]

Step 1: Preprocess arr2

• Sort arr2: [1, 3, 4]
• Remove duplicates: [1, 3, 4] (no duplicates in this case)

Step 2: Add sentinels

• arr = [-inf, 1, 5, 3, 6, inf]
• Length n = 6

Step 3: Initialize DP array

• f = [0, inf, inf, inf, inf, inf]
• f[0] = 0 (no operations needed for sentinel)

Step 4: Fill DP array

Position i = 1 (value = 1):

• Check if arr[0] < arr[1]: -inf < 1 ✓
• Set f[1] = f[0] = 0 (can keep arr[1] = 1)
• Binary search: j = bisect_left([1,3,4], 1) = 0
• No replacements possible (j = 0)
• Result: f[1] = 0

Position i = 2 (value = 5):

• Check if arr[1] < arr[2]: 1 < 5 ✓
• Set f[2] = f[1] = 0 (can keep arr[2] = 5)
• Binary search: j = bisect_left([1,3,4], 5) = 3
• Try k = 1: Replace 1 element before position 2
  • Need arr[0] < arr2[2]: -inf < 4 ✓
  • Update: f[2] = min(0, f[0] + 1) = 0
• Result: f[2] = 0

Position i = 3 (value = 3):

• Check if arr[2] < arr[3]: 5 < 3 ✗ (cannot directly keep both)
• Binary search: j = bisect_left([1,3,4], 3) = 1
• Try k = 1: Replace 1 element before position 3 (replace position 2)
  • Need arr[1] < arr2[0]: 1 < 1 ✗ (not valid)
• Result: f[3] = inf (cannot keep value 3 at position 3)

Position i = 4 (value = 6):

• Check if arr[3] < arr[4]: 3 < 6 ✓
• Set f[4] = f[3] = inf (but f[3] is inf)
• Binary search: j = bisect_left([1,3,4], 6) = 3
• Try k = 1: Replace 1 element before position 4 (replace position 3)
  • Need arr[2] < arr2[2]: 5 < 4 ✗ (not valid)
• Try k = 2: Replace 2 elements before position 4 (replace positions 2 and 3)
  • Need arr[1] < arr2[1]: 1 < 3 ✓
  • Update: f[4] = min(inf, f[1] + 2) = 0 + 2 = 2
• Try k = 3: Replace 3 elements before position 4 (replace positions 1, 2, and 3)
  • Need arr[0] < arr2[0]: -inf < 1 ✓
  • Update: f[4] = min(2, f[0] + 3) = min(2, 3) = 2
• Result: f[4] = 2

Position i = 5 (value = inf):

• Check if arr[4] < arr[5]: 6 < inf ✓
• Set f[5] = f[4] = 2
• Result: f[5] = 2

Step 5: Return result

• f[5] = 2, which is less than inf
• Return 2

The minimum number of operations is 2. We need to replace positions 2 and 3 in the original array:

• Replace arr1[1] = 5 with arr2[1] = 3
• Replace arr1[2] = 3 with arr2[2] = 4
• Final array: [1, 3, 4, 6] which is strictly increasing

Solution Implementation

Time and Space Complexity

Time Complexity: O(n × (log m + min(m, n)))

The algorithm consists of several key operations:

• Sorting arr2 takes O(m log m) where m is the length of arr2
• Removing duplicates from arr2 takes O(m) time
• The main dynamic programming loop runs for n iterations (length of augmented arr)
• For each position i:
  • Checking if arr[i-1] < arr[i] takes O(1)
  • Binary search using bisect_left takes O(log m)
  • The inner loop runs for min(i-1, j) iterations, which is bounded by min(m, n)
  • Each iteration of the inner loop performs constant time operations

Therefore, the dominant operation is the nested loop structure with binary search, giving us O(n × (log m + min(m, n))) time complexity.

Space Complexity: O(n)

The space usage includes:

• The modified arr2 after removing duplicates uses O(m) space, but this modifies the input in-place
• The augmented array arr uses O(n) space
• The DP array f uses O(n) space
• Other variables use O(1) space

Since we're measuring additional space beyond the input, and n is the length of arr1, the space complexity is O(n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Not Handling Duplicate Values in arr2

Pitfall: Using arr2 directly without removing duplicates can lead to inefficient binary search and incorrect replacement logic. When duplicates exist, the algorithm might consider the same value multiple times for replacement, which doesn't provide any benefit but complicates the logic.

Example:

arr1 = [1, 5, 3]
arr2 = [2, 2, 2, 4, 4]  # Contains duplicates

Solution: Always sort and deduplicate arr2 first:

arr2.sort()
unique_count = 0
for value in arr2:
    if unique_count == 0 or value != arr2[unique_count - 1]:
        arr2[unique_count] = value
        unique_count += 1
arr2 = arr2[:unique_count]

2. Incorrect Boundary Checking When Replacing Multiple Elements

Pitfall: When trying to replace k consecutive elements, failing to properly check if there are enough elements available in both arrays can cause index out of bounds errors or incorrect results.

Wrong approach:

for k in range(1, i):  # Could exceed available elements in arr2
    if extended_arr[i - k - 1] < arr2[max_replacement_idx - k]:
        # This might access arr2[-1] when max_replacement_idx < k

Solution: Use min(i - 1, max_replacement_idx) to ensure we don't exceed bounds:

for k in range(1, min(i - 1, max_replacement_idx) + 1):
    if extended_arr[i - k - 1] < arr2[max_replacement_idx - k]:
        dp[i] = min(dp[i], dp[i - k - 1] + k)

3. Forgetting to Check if Keeping Current Element is Valid

Pitfall: Always trying to replace elements without first checking if the current configuration already works.

Example:

# Wrong: Only considering replacements
for i in range(1, n):
    # Only replacement logic, missing the "keep current" option
    max_replacement_idx = bisect_left(arr2, extended_arr[i])
    for k in range(1, min(i - 1, max_replacement_idx) + 1):
        ...

Solution: Always check if keeping the current element maintains strict increasing order:

if extended_arr[i - 1] < extended_arr[i]:
    dp[i] = dp[i - 1]  # Can keep current element without replacement

4. Using Regular Infinity Instead of Float Infinity

Pitfall: Using a large integer like 10**9 as infinity can cause issues if the actual minimum operations exceed this value or when comparing with legitimate values.

Wrong:

dp = [10**9] * n  # Might not be large enough
extended_arr = [-10**9] + arr1 + [10**9]  # Might conflict with actual values

Solution: Use float('inf') and float('-inf'):

dp = [float('inf')] * n
extended_arr = [float('-inf')] + arr1 + [float('inf')]

5. Misunderstanding the Replacement Strategy

Pitfall: Thinking that we need to find the optimal replacement for each position independently, rather than considering consecutive replacements as a group.

Key Insight: When replacing k consecutive elements before position i, we should use the largest k values from arr2 that are less than extended_arr[i]. These are arr2[max_replacement_idx-k:max_replacement_idx], which ensures:

• They form a strictly increasing sequence
• They're all less than extended_arr[i]
• They're the largest possible values satisfying these conditions (minimizing conflict with earlier elements)

Solution: The code correctly handles this by:

1. Finding max_replacement_idx using binary search
2. Taking k elements ending at max_replacement_idx-1
3. Checking if the smallest of these (arr2[max_replacement_idx-k]) is greater than extended_arr[i-k-1]