LeetCode #1827 — EASY

Minimum Operations to Make the Array Increasing

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given an integer array nums (0-indexed). In one operation, you can choose an element of the array and increment it by 1.

For example, if nums = [1,2,3], you can choose to increment nums[1] to make nums = [1,3,3].

Return the minimum number of operations needed to make nums strictly increasing.

An array nums is strictly increasing if nums[i] < nums[i+1] for all 0 <= i < nums.length - 1. An array of length 1 is trivially strictly increasing.

Example 1:

Input: nums = [1,1,1]
Output: 3
Explanation: You can do the following operations:
1) Increment nums[2], so nums becomes [1,1,2].
2) Increment nums[1], so nums becomes [1,2,2].
3) Increment nums[2], so nums becomes [1,2,3].

Example 2:

Input: nums = [1,5,2,4,1]
Output: 14

Example 3:

Input: nums = [8]
Output: 0

Constraints:

1 <= nums.length <= 5000
1 <= nums[i] <= 10⁴

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums (0-indexed). In one operation, you can choose an element of the array and increment it by 1. For example, if nums = [1,2,3], you can choose to increment nums[1] to make nums = [1,3,3]. Return the minimum number of operations needed to make nums strictly increasing. An array nums is strictly increasing if nums[i] < nums[i+1] for all 0 <= i < nums.length - 1. An array of length 1 is trivially strictly increasing.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[1,1,1]

Example 2

[1,5,2,4,1]

Example 3

[8]

Core Insight

What unlocks the optimal approach

nums[i+1] must be at least equal to nums[i] + 1.
Think greedily. You don't have to increase nums[i+1] beyond nums[i]+1.
Iterate on i and set nums[i] = max(nums[i-1]+1, nums[i]) .

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1827: Minimum Operations to Make the Array Increasing
class Solution {
    public int minOperations(int[] nums) {
        int ans = 0, mx = 0;
        for (int v : nums) {
            ans += Math.max(0, mx + 1 - v);
            mx = Math.max(mx + 1, v);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1827: Minimum Operations to Make the Array Increasing
func minOperations(nums []int) (ans int) {
	mx := 0
	for _, v := range nums {
		ans += max(0, mx+1-v)
		mx = max(mx+1, v)
	}
	return
}

# Accepted solution for LeetCode #1827: Minimum Operations to Make the Array Increasing
class Solution:
    def minOperations(self, nums: List[int]) -> int:
        ans = mx = 0
        for v in nums:
            ans += max(0, mx + 1 - v)
            mx = max(mx + 1, v)
        return ans

// Accepted solution for LeetCode #1827: Minimum Operations to Make the Array Increasing
impl Solution {
    pub fn min_operations(nums: Vec<i32>) -> i32 {
        let mut ans = 0;
        let mut max = 0;
        for &v in nums.iter() {
            ans += (0).max(max + 1 - v);
            max = v.max(max + 1);
        }
        ans
    }
}

// Accepted solution for LeetCode #1827: Minimum Operations to Make the Array Increasing
function minOperations(nums: number[]): number {
    let ans = 0;
    let max = 0;
    for (const v of nums) {
        ans += Math.max(0, max + 1 - v);
        max = Math.max(max + 1, v);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.

Using greedy without proof

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.