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 an integer array target. You have an integer array initial of the same size as target with all elements initially zeros.
In one operation you can choose any subarray from initial and increment each value by one.
Return the minimum number of operations to form a target array from initial.
The test cases are generated so that the answer fits in a 32-bit integer.
Example 1:
Input: target = [1,2,3,2,1] Output: 3 Explanation: We need at least 3 operations to form the target array from the initial array. [0,0,0,0,0] increment 1 from index 0 to 4 (inclusive). [1,1,1,1,1] increment 1 from index 1 to 3 (inclusive). [1,2,2,2,1] increment 1 at index 2. [1,2,3,2,1] target array is formed.
Example 2:
Input: target = [3,1,1,2] Output: 4 Explanation: [0,0,0,0] -> [1,1,1,1] -> [1,1,1,2] -> [2,1,1,2] -> [3,1,1,2]
Example 3:
Input: target = [3,1,5,4,2] Output: 7 Explanation: [0,0,0,0,0] -> [1,1,1,1,1] -> [2,1,1,1,1] -> [3,1,1,1,1] -> [3,1,2,2,2] -> [3,1,3,3,2] -> [3,1,4,4,2] -> [3,1,5,4,2].
Constraints:
1 <= target.length <= 1051 <= target[i] <= 105Problem summary: You are given an integer array target. You have an integer array initial of the same size as target with all elements initially zeros. In one operation you can choose any subarray from initial and increment each value by one. Return the minimum number of operations to form a target array from initial. The test cases are generated so that the answer fits in a 32-bit integer.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Dynamic Programming · Stack · Greedy
[1,2,3,2,1]
[3,1,1,2]
[3,1,5,4,2]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1526: Minimum Number of Increments on Subarrays to Form a Target Array
class Solution {
public int minNumberOperations(int[] target) {
int f = target[0];
for (int i = 1; i < target.length; ++i) {
if (target[i] > target[i - 1]) {
f += target[i] - target[i - 1];
}
}
return f;
}
}
// Accepted solution for LeetCode #1526: Minimum Number of Increments on Subarrays to Form a Target Array
func minNumberOperations(target []int) int {
f := target[0]
for i, x := range target[1:] {
if x > target[i] {
f += x - target[i]
}
}
return f
}
# Accepted solution for LeetCode #1526: Minimum Number of Increments on Subarrays to Form a Target Array
class Solution:
def minNumberOperations(self, target: List[int]) -> int:
return target[0] + sum(max(0, b - a) for a, b in pairwise(target))
// Accepted solution for LeetCode #1526: Minimum Number of Increments on Subarrays to Form a Target Array
impl Solution {
pub fn min_number_operations(target: Vec<i32>) -> i32 {
let mut f = target[0];
for i in 1..target.len() {
if target[i] > target[i - 1] {
f += target[i] - target[i - 1];
}
}
f
}
}
// Accepted solution for LeetCode #1526: Minimum Number of Increments on Subarrays to Form a Target Array
function minNumberOperations(target: number[]): number {
let f = target[0];
for (let i = 1; i < target.length; ++i) {
if (target[i] > target[i - 1]) {
f += target[i] - target[i - 1];
}
}
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.