LeetCode #3048 — MEDIUM

Earliest Second to Mark Indices I

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given two 1-indexed integer arrays, nums and, changeIndices, having lengths n and m, respectively.

Initially, all indices in nums are unmarked. Your task is to mark all indices in nums.

In each second, s, in order from 1 to m (inclusive), you can perform one of the following operations:

Choose an index i in the range [1, n] and decrement nums[i] by 1.
If nums[changeIndices[s]] is equal to 0, mark the index changeIndices[s].
Do nothing.

Return an integer denoting the earliest second in the range [1, m] when all indices in nums can be marked by choosing operations optimally, or -1 if it is impossible.

Example 1:

Input: nums = [2,2,0], changeIndices = [2,2,2,2,3,2,2,1]
Output: 8
Explanation: In this example, we have 8 seconds. The following operations can be performed to mark all indices:
Second 1: Choose index 1 and decrement nums[1] by one. nums becomes [1,2,0].
Second 2: Choose index 1 and decrement nums[1] by one. nums becomes [0,2,0].
Second 3: Choose index 2 and decrement nums[2] by one. nums becomes [0,1,0].
Second 4: Choose index 2 and decrement nums[2] by one. nums becomes [0,0,0].
Second 5: Mark the index changeIndices[5], which is marking index 3, since nums[3] is equal to 0.
Second 6: Mark the index changeIndices[6], which is marking index 2, since nums[2] is equal to 0.
Second 7: Do nothing.
Second 8: Mark the index changeIndices[8], which is marking index 1, since nums[1] is equal to 0.
Now all indices have been marked.
It can be shown that it is not possible to mark all indices earlier than the 8th second.
Hence, the answer is 8.

Example 2:

Input: nums = [1,3], changeIndices = [1,1,1,2,1,1,1]
Output: 6
Explanation: In this example, we have 7 seconds. The following operations can be performed to mark all indices:
Second 1: Choose index 2 and decrement nums[2] by one. nums becomes [1,2].
Second 2: Choose index 2 and decrement nums[2] by one. nums becomes [1,1].
Second 3: Choose index 2 and decrement nums[2] by one. nums becomes [1,0].
Second 4: Mark the index changeIndices[4], which is marking index 2, since nums[2] is equal to 0.
Second 5: Choose index 1 and decrement nums[1] by one. nums becomes [0,0].
Second 6: Mark the index changeIndices[6], which is marking index 1, since nums[1] is equal to 0.
Now all indices have been marked.
It can be shown that it is not possible to mark all indices earlier than the 6th second.
Hence, the answer is 6.

Example 3:

Input: nums = [0,1], changeIndices = [2,2,2]
Output: -1
Explanation: In this example, it is impossible to mark all indices because index 1 isn't in changeIndices.
Hence, the answer is -1.

Constraints:

1 <= n == nums.length <= 2000
0 <= nums[i] <= 10⁹
1 <= m == changeIndices.length <= 2000
1 <= changeIndices[i] <= n

Step 01

Brute Force Baseline

Problem summary: You are given two 1-indexed integer arrays, nums and, changeIndices, having lengths n and m, respectively. Initially, all indices in nums are unmarked. Your task is to mark all indices in nums. In each second, s, in order from 1 to m (inclusive), you can perform one of the following operations: Choose an index i in the range [1, n] and decrement nums[i] by 1. If nums[changeIndices[s]] is equal to 0, mark the index changeIndices[s]. Do nothing. Return an integer denoting the earliest second in the range [1, m] when all indices in nums can be marked by choosing operations optimally, or -1 if it is impossible.

Baseline thinking

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

Pattern signal: Array · Binary Search

Example 1

[2,2,0]
[2,2,2,2,3,2,2,1]

Example 2

[1,3]
[1,1,1,2,1,1,1]

Example 3

[0,1]
[2,2,2]

Step 02

Core Insight

What unlocks the optimal approach

Consider using binary search.
Suppose the <code>answer <= x</code>; we can mark each index as late as possible. Namely, mark each index at the last occurrence in the array <code>changeIndices[1..x]</code>.
When marking an index, which is the last occurrence at the second <code>i</code>, we check whether we have a sufficient number of decrement operations to mark all the previous indices whose last occurrences have already been marked, and the current index, i.e., <code>i - sum_of_marked_indices_values - cnt_of_marked_indices >= nums[changeIndices[i]]</code>.
The answer is the earliest second when all indices can be marked after running the binary search or <code>-1</code> if there is no such second.

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 #3048: Earliest Second to Mark Indices I
class Solution {
    private int[] nums;
    private int[] changeIndices;

    public int earliestSecondToMarkIndices(int[] nums, int[] changeIndices) {
        this.nums = nums;
        this.changeIndices = changeIndices;
        int m = changeIndices.length;
        int l = 1, r = m + 1;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (check(mid)) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l > m ? -1 : l;
    }

    private boolean check(int t) {
        int[] last = new int[nums.length + 1];
        for (int s = 0; s < t; ++s) {
            last[changeIndices[s]] = s;
        }
        int decrement = 0;
        int marked = 0;
        for (int s = 0; s < t; ++s) {
            int i = changeIndices[s];
            if (last[i] == s) {
                if (decrement < nums[i - 1]) {
                    return false;
                }
                decrement -= nums[i - 1];
                ++marked;
            } else {
                ++decrement;
            }
        }
        return marked == nums.length;
    }
}

// Accepted solution for LeetCode #3048: Earliest Second to Mark Indices I
func earliestSecondToMarkIndices(nums []int, changeIndices []int) int {
	n, m := len(nums), len(changeIndices)
	l := sort.Search(m+1, func(t int) bool {
		last := make([]int, n+1)
		for s, i := range changeIndices[:t] {
			last[i] = s
		}
		decrement, marked := 0, 0
		for s, i := range changeIndices[:t] {
			if last[i] == s {
				if decrement < nums[i-1] {
					return false
				}
				decrement -= nums[i-1]
				marked++
			} else {
				decrement++
			}
		}
		return marked == n
	})
	if l > m {
		return -1
	}
	return l
}

# Accepted solution for LeetCode #3048: Earliest Second to Mark Indices I
class Solution:
    def earliestSecondToMarkIndices(
        self, nums: List[int], changeIndices: List[int]
    ) -> int:
        def check(t: int) -> bool:
            decrement = 0
            marked = 0
            last = {i: s for s, i in enumerate(changeIndices[:t])}
            for s, i in enumerate(changeIndices[:t]):
                if last[i] == s:
                    if decrement < nums[i - 1]:
                        return False
                    decrement -= nums[i - 1]
                    marked += 1
                else:
                    decrement += 1
            return marked == len(nums)

        m = len(changeIndices)
        l = bisect_left(range(1, m + 2), True, key=check) + 1
        return -1 if l > m else l

// Accepted solution for LeetCode #3048: Earliest Second to Mark Indices I
// 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 #3048: Earliest Second to Mark Indices I
// class Solution {
//     private int[] nums;
//     private int[] changeIndices;
// 
//     public int earliestSecondToMarkIndices(int[] nums, int[] changeIndices) {
//         this.nums = nums;
//         this.changeIndices = changeIndices;
//         int m = changeIndices.length;
//         int l = 1, r = m + 1;
//         while (l < r) {
//             int mid = (l + r) >> 1;
//             if (check(mid)) {
//                 r = mid;
//             } else {
//                 l = mid + 1;
//             }
//         }
//         return l > m ? -1 : l;
//     }
// 
//     private boolean check(int t) {
//         int[] last = new int[nums.length + 1];
//         for (int s = 0; s < t; ++s) {
//             last[changeIndices[s]] = s;
//         }
//         int decrement = 0;
//         int marked = 0;
//         for (int s = 0; s < t; ++s) {
//             int i = changeIndices[s];
//             if (last[i] == s) {
//                 if (decrement < nums[i - 1]) {
//                     return false;
//                 }
//                 decrement -= nums[i - 1];
//                 ++marked;
//             } else {
//                 ++decrement;
//             }
//         }
//         return marked == nums.length;
//     }
// }

// Accepted solution for LeetCode #3048: Earliest Second to Mark Indices I
function earliestSecondToMarkIndices(nums: number[], changeIndices: number[]): number {
    const [n, m] = [nums.length, changeIndices.length];
    let [l, r] = [1, m + 1];
    const check = (t: number): boolean => {
        const last: number[] = Array(n + 1).fill(0);
        for (let s = 0; s < t; ++s) {
            last[changeIndices[s]] = s;
        }
        let [decrement, marked] = [0, 0];
        for (let s = 0; s < t; ++s) {
            const i = changeIndices[s];
            if (last[i] === s) {
                if (decrement < nums[i - 1]) {
                    return false;
                }
                decrement -= nums[i - 1];
                ++marked;
            } else {
                ++decrement;
            }
        }
        return marked === n;
    };
    while (l < r) {
        const mid = (l + r) >> 1;
        if (check(mid)) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return l > m ? -1 : l;
}

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(m × log m)

Space

O(n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.