Problem Description

You have two integer arrays nums1 and nums2 that are both the same length n, and an integer k.

You can perform the following operation on nums1:

• Pick any two different indices i and j
• Add k to nums1[i]
• Subtract k from nums1[j]

This operation maintains the total sum of nums1 since you're adding k to one element and subtracting k from another.

Your goal is to make nums1 exactly equal to nums2, meaning every element at each index must match: nums1[i] == nums2[i] for all indices from 0 to n-1.

Return the minimum number of operations needed to achieve this. If it's impossible to make the arrays equal, return -1.

For example, if nums1 = [4, 3, 1, 4], nums2 = [1, 3, 7, 1], and k = 3:

• You could add 3 to nums1[0] and subtract 3 from nums1[2], making nums1 = [7, 3, -2, 4]
• Then add 3 to nums1[2] twice and subtract 3 from nums1[3] twice, eventually reaching nums1 = [1, 3, 7, 1]

The solution needs to handle edge cases like when k = 0 (no changes possible unless arrays are already equal) and when the differences between corresponding elements aren't divisible by k (making transformation impossible).

Intuition

Let's think about what each operation actually does. When we add k to nums1[i] and subtract k from nums1[j], we're essentially transferring k units from position j to position i. The total sum of the array remains unchanged.

For each index i, we need nums1[i] to become nums2[i]. The difference nums1[i] - nums2[i] tells us how much excess or deficit we have at position i. If this difference is positive, we have excess that needs to be removed. If negative, we have a deficit that needs to be filled.

Key observations:

1. Since each operation transfers exactly k units, any difference nums1[i] - nums2[i] must be divisible by k. If not, it's impossible to fix that position.

2. The total sum of nums1 never changes through our operations. Therefore, if sum(nums1) != sum(nums2), it's impossible to make them equal.

3. Every unit of excess at one position must be matched with a unit of deficit at another position. If we have +3k excess at position i, we need exactly -3k deficit somewhere else to balance it out.

4. To transfer m*k units from position j to position i, we need exactly |m| operations. But here's the clever part: when we count the total operations needed, we're counting each transfer twice - once as removing excess from position j and once as filling deficit at position i. That's why the final answer is total_transfers / 2.

The special case k = 0 means no transfers are possible, so the arrays must already be equal at every position.

Learn more about Greedy and Math patterns.

Solution Approach

The implementation uses a single-pass algorithm to check feasibility and calculate the minimum operations simultaneously.

We maintain two key variables:

• x: Tracks the cumulative difference between nums1 and nums2. This helps verify if the total sum matches.
• ans: Accumulates the total absolute differences, which represents the total amount of transfers needed.

Algorithm Steps:

1. Iterate through both arrays simultaneously using zip(nums1, nums2) to process corresponding elements.

2. Handle the special case k = 0:

   • If k = 0, no transfers are possible
   • If a != b at any position, return -1 immediately
   • Otherwise, continue to the next position

3. Check divisibility for each difference:

   • Calculate (a - b) for current position
   • If (a - b) % k != 0, the difference cannot be resolved with transfers of size k, so return -1

4. Track the transfers:

   • Calculate y = (a - b) // k, which represents how many transfers of size k are needed
   • Add abs(y) to ans to count the total transfer amount
   • Add y (with sign) to x to track the net difference

5. Final validation:

   • After processing all positions, check if x == 0
   • If x != 0, the sums don't match, making transformation impossible, so return -1
   • Otherwise, return ans // 2

Why ans // 2? Each transfer involves two positions - one giving and one receiving. When we sum up abs(y) for all positions, we count each transfer twice: once as a positive difference (excess) and once as a negative difference (deficit). Therefore, the actual number of operations is half of the total absolute differences.

Time Complexity: O(n) where n is the length of the arrays Space Complexity: O(1) as we only use a constant amount of extra space

Example Walkthrough

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

Example: nums1 = [3, 8, 5, 2], nums2 = [7, 2, 3, 6], k = 2

Step 1: Calculate differences at each position

• Position 0: 3 - 7 = -4 (need to add 4)
• Position 1: 8 - 2 = 6 (need to remove 6)
• Position 2: 5 - 3 = 2 (need to remove 2)
• Position 3: 2 - 6 = -4 (need to add 4)

Step 2: Check feasibility

• All differences (-4, 6, 2, -4) are divisible by k = 2 ✓
• Sum of differences: -4 + 6 + 2 + (-4) = 0 ✓
• Total sum check: sum(nums1) = 18, sum(nums2) = 18 ✓

Step 3: Calculate transfers needed

• Position 0: -4 ÷ 2 = -2 transfers (needs 2 incoming transfers)
• Position 1: 6 ÷ 2 = 3 transfers (needs 3 outgoing transfers)
• Position 2: 2 ÷ 2 = 1 transfer (needs 1 outgoing transfer)
• Position 3: -4 ÷ 2 = -2 transfers (needs 2 incoming transfers)

Step 4: Count total transfer amount

• Total absolute transfers: |-2| + |3| + |1| + |-2| = 2 + 3 + 1 + 2 = 8

Step 5: Calculate operations

• Since each operation moves k units from one position to another, and we counted each transfer twice (once as outgoing, once as incoming), the answer is 8 ÷ 2 = 4 operations.

Verification of the operations:

1. Transfer from position 1 to position 0: nums1 = [5, 6, 5, 2]
2. Transfer from position 1 to position 0: nums1 = [7, 4, 5, 2]
3. Transfer from position 1 to position 3: nums1 = [7, 2, 5, 4]
4. Transfer from position 2 to position 3: nums1 = [7, 2, 3, 6]

After 4 operations, nums1 equals nums2!

Solution Implementation

Time and Space Complexity

The time complexity of this solution is O(n), where n is the length of the arrays nums1 and nums2. This is because the algorithm iterates through both arrays exactly once using zip(nums1, nums2), performing constant-time operations for each pair of elements (arithmetic operations, modulo checks, and variable updates).

The space complexity is O(1) as the algorithm only uses a fixed amount of extra space regardless of the input size. The variables ans, x, a, b, and y are the only additional storage used, and their space requirements don't scale with the input size n.

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

Common Pitfalls

1. Forgetting to Handle the k = 0 Special Case

Pitfall: When k = 0, no operations are possible since adding/subtracting 0 doesn't change any values. Many solutions forget this edge case and might attempt division by zero or return incorrect results.

Example of incorrect handling:

# WRONG - This will cause ZeroDivisionError
operations_at_index = (num1_val - num2_val) // k  # Crashes when k = 0

Solution: Always check for k = 0 first and handle it separately:

if k == 0:
    if num1_val != num2_val:
        return -1  # Can't make arrays equal
    continue  # Skip to next element if already equal

2. Not Checking if the Sum of Arrays Match

Pitfall: Since each operation adds k to one element and subtracts k from another, the total sum of nums1 must equal the total sum of nums2 for transformation to be possible. Failing to verify this leads to incorrect results.

Example scenario that would fail without sum checking:

• nums1 = [1, 2, 3], nums2 = [2, 3, 5], k = 1
• Sum of nums1 = 6, Sum of nums2 = 10
• Without the net_change check, the algorithm might not detect this impossibility

Solution: Track the net change throughout the iteration and verify it equals zero at the end:

net_change += operations_at_index
# After loop
if net_change != 0:
    return -1

3. Forgetting to Divide the Total Operations by 2

Pitfall: Each operation involves two indices - one gains k and one loses k. When summing absolute differences, each operation gets counted twice (once as positive difference, once as negative).

Example of the double-counting issue:

• nums1 = [5, 3], nums2 = [3, 5], k = 2
• Index 0 needs -1 operation (subtract 2), Index 1 needs +1 operation (add 2)
• Total absolute differences = |−1| + |1| = 2
• But this represents just ONE operation (moving 2 from index 0 to index 1)

Solution: Always divide the final count by 2:

return total_operations // 2  # Not just total_operations

4. Using Modulo Check Incorrectly with Negative Numbers

Pitfall: In Python, modulo with negative numbers works correctly, but conceptually some might check divisibility incorrectly or use absolute values unnecessarily.

Incorrect approach:

# WRONG - Using absolute value loses important sign information
if abs(num1_val - num2_val) % k != 0:
    return -1
operations = abs(num1_val - num2_val) // k  # Lost direction of transfer

Solution: Keep the sign information intact:

if (num1_val - num2_val) % k != 0:
    return -1
operations_at_index = (num1_val - num2_val) // k  # Preserves sign for net_change tracking