Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
You are given two positive integer arrays nums and target, of the same length.
In a single operation, you can select any subarray of nums and increment each element within that subarray by 1 or decrement each element within that subarray by 1.
Return the minimum number of operations required to make nums equal to the array target.
Example 1:
Input: nums = [3,5,1,2], target = [4,6,2,4]
Output: 2
Explanation:
We will perform the following operations to make nums equal to target:
- Increment nums[0..3] by 1, nums = [4,6,2,3].
- Increment nums[3..3] by 1, nums = [4,6,2,4].
Example 2:
Input: nums = [1,3,2], target = [2,1,4]
Output: 5
Explanation:
We will perform the following operations to make nums equal to target:
- Increment nums[0..0] by 1, nums = [2,3,2].
- Decrement nums[1..1] by 1, nums = [2,2,2].
- Decrement nums[1..1] by 1, nums = [2,1,2].
- Increment nums[2..2] by 1, nums = [2,1,3].
- Increment nums[2..2] by 1, nums = [2,1,4].
Constraints:
1 <= nums.length == target.length <= 1051 <= nums[i], target[i] <= 108Problem summary: You are given two positive integer arrays nums and target, of the same length. In a single operation, you can select any subarray of nums and increment each element within that subarray by 1 or decrement each element within that subarray by 1. Return the minimum number of operations required to make nums equal to the array target.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Dynamic Programming · Stack · Greedy
[3,5,1,2] [4,6,2,4]
[1,3,2] [2,1,4]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3229: Minimum Operations to Make Array Equal to Target
class Solution {
public long minimumOperations(int[] nums, int[] target) {
long f = Math.abs(target[0] - nums[0]);
for (int i = 1; i < nums.length; ++i) {
long x = target[i] - nums[i];
long y = target[i - 1] - nums[i - 1];
if (x * y > 0) {
long d = Math.abs(x) - Math.abs(y);
if (d > 0) {
f += d;
}
} else {
f += Math.abs(x);
}
}
return f;
}
}
// Accepted solution for LeetCode #3229: Minimum Operations to Make Array Equal to Target
func minimumOperations(nums []int, target []int) int64 {
f := abs(target[0] - nums[0])
for i := 1; i < len(target); i++ {
x := target[i] - nums[i]
y := target[i-1] - nums[i-1]
if x*y > 0 {
if d := abs(x) - abs(y); d > 0 {
f += d
}
} else {
f += abs(x)
}
}
return int64(f)
}
func abs(x int) int {
if x < 0 {
return -x
}
return x
}
# Accepted solution for LeetCode #3229: Minimum Operations to Make Array Equal to Target
class Solution:
def minimumOperations(self, nums: List[int], target: List[int]) -> int:
n = len(nums)
f = abs(target[0] - nums[0])
for i in range(1, n):
x = target[i] - nums[i]
y = target[i - 1] - nums[i - 1]
if x * y > 0:
d = abs(x) - abs(y)
if d > 0:
f += d
else:
f += abs(x)
return f
// Accepted solution for LeetCode #3229: Minimum Operations to Make Array Equal to Target
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
pub fn rust_example() {
// Port the logic from the reference block below.
}
}
// Reference (java):
// // Accepted solution for LeetCode #3229: Minimum Operations to Make Array Equal to Target
// class Solution {
// public long minimumOperations(int[] nums, int[] target) {
// long f = Math.abs(target[0] - nums[0]);
// for (int i = 1; i < nums.length; ++i) {
// long x = target[i] - nums[i];
// long y = target[i - 1] - nums[i - 1];
// if (x * y > 0) {
// long d = Math.abs(x) - Math.abs(y);
// if (d > 0) {
// f += d;
// }
// } else {
// f += Math.abs(x);
// }
// }
// return f;
// }
// }
// Accepted solution for LeetCode #3229: Minimum Operations to Make Array Equal to Target
function minimumOperations(nums: number[], target: number[]): number {
const n = nums.length;
let f = Math.abs(target[0] - nums[0]);
for (let i = 1; i < n; ++i) {
const x = target[i] - nums[i];
const y = target[i - 1] - nums[i - 1];
if (x * y > 0) {
const d = Math.abs(x) - Math.abs(y);
if (d > 0) {
f += d;
}
} else {
f += Math.abs(x);
}
}
return f;
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.
Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.
Usually fails on: Indices point to blocked elements and outputs shift.
Fix: Pop while invariant is violated before pushing current element.
Wrong move: Locally optimal choices may fail globally.
Usually fails on: Counterexamples appear on crafted input orderings.
Fix: Verify with exchange argument or monotonic objective before committing.