LeetCode #3468 — MEDIUM

Find the Number of Copy Arrays

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

The Problem

Problem Statement

You are given an array original of length n and a 2D array bounds of length n x 2, where bounds[i] = [u_i, v_i].

You need to find the number of possible arrays copy of length n such that:

(copy[i] - copy[i - 1]) == (original[i] - original[i - 1]) for 1 <= i <= n - 1.
u_i <= copy[i] <= v_i for 0 <= i <= n - 1.

Return the number of such arrays.

Example 1:

Input: original = [1,2,3,4], bounds = [[1,2],[2,3],[3,4],[4,5]]

Output: 2

Explanation:

The possible arrays are:

[1, 2, 3, 4]
[2, 3, 4, 5]

Example 2:

Input: original = [1,2,3,4], bounds = [[1,10],[2,9],[3,8],[4,7]]

Output: 4

Explanation:

The possible arrays are:

[1, 2, 3, 4]
[2, 3, 4, 5]
[3, 4, 5, 6]
[4, 5, 6, 7]

Example 3:

Input: original = [1,2,1,2], bounds = [[1,1],[2,3],[3,3],[2,3]]

Output: 0

Explanation:

No array is possible.

Constraints:

2 <= n == original.length <= 10⁵
1 <= original[i] <= 10⁹
bounds.length == n
bounds[i].length == 2
1 <= bounds[i][0] <= bounds[i][1] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given an array original of length n and a 2D array bounds of length n x 2, where bounds[i] = [ui, vi]. You need to find the number of possible arrays copy of length n such that: (copy[i] - copy[i - 1]) == (original[i] - original[i - 1]) for 1 <= i <= n - 1. ui <= copy[i] <= vi for 0 <= i <= n - 1. Return the number of such arrays.

Baseline thinking

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

Pattern signal: Array · Math

Example 1

[1,2,3,4]
[[1,2],[2,3],[3,4],[4,5]]

Example 2

[1,2,3,4]
[[1,10],[2,9],[3,8],[4,7]]

Example 3

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

Core Insight

What unlocks the optimal approach

<code>copy[0]</code> uniquely determines all other values.
Possible values for <code>copy[0]</code> are in <code>[u[0], v[0]]</code>.
From left to right, compute valid ranges for each index by intersecting bounds with the previous range.
The answer is the size of the valid range for the last index.

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 #3468: Find the Number of Copy Arrays
class Solution {
  public int countArrays(int[] original, int[][] bounds) {
    int mn = bounds[0][0];
    int mx = bounds[0][1];

    for (int i = 1; i < original.length; ++i) {
      final int diff = original[i] - original[i - 1];
      mn = Math.max(mn + diff, bounds[i][0]);
      mx = Math.min(mx + diff, bounds[i][1]);
    }

    return Math.max(0, mx - mn + 1);
  }
}

// Accepted solution for LeetCode #3468: Find the Number of Copy Arrays
package main

import "math"

// https://space.bilibili.com/206214
func countArrays(original []int, bounds [][]int) int {
	mn, mx := math.MinInt, math.MaxInt
	for i, b := range bounds {
		mn = max(mn, b[0]-original[i]) // 计算区间交集
		mx = min(mx, b[1]-original[i])
	}
	return max(mx-mn+1, 0) // 注意交集可能是空的
}

# Accepted solution for LeetCode #3468: Find the Number of Copy Arrays
class Solution:
  def countArrays(self, original: list[int], bounds: list[list[int]]) -> int:
    mn, mx = bounds[0]

    for i in range(1, len(original)):
      diff = original[i] - original[i - 1]
      mn = max(mn + diff, bounds[i][0])
      mx = min(mx + diff, bounds[i][1])

    return max(0, mx - mn + 1)

// Accepted solution for LeetCode #3468: Find the Number of Copy Arrays
// 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 #3468: Find the Number of Copy Arrays
// class Solution {
//   public int countArrays(int[] original, int[][] bounds) {
//     int mn = bounds[0][0];
//     int mx = bounds[0][1];
// 
//     for (int i = 1; i < original.length; ++i) {
//       final int diff = original[i] - original[i - 1];
//       mn = Math.max(mn + diff, bounds[i][0]);
//       mx = Math.min(mx + diff, bounds[i][1]);
//     }
// 
//     return Math.max(0, mx - mn + 1);
//   }
// }

// Accepted solution for LeetCode #3468: Find the Number of Copy Arrays
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #3468: Find the Number of Copy Arrays
// class Solution {
//   public int countArrays(int[] original, int[][] bounds) {
//     int mn = bounds[0][0];
//     int mx = bounds[0][1];
// 
//     for (int i = 1; i < original.length; ++i) {
//       final int diff = original[i] - original[i - 1];
//       mn = Math.max(mn + diff, bounds[i][0]);
//       mx = Math.min(mx + diff, bounds[i][1]);
//     }
// 
//     return Math.max(0, mx - mn + 1);
//   }
// }

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)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Find the Number of Copy Arrays

Problem Statement

Roadmap

Brute Force Baseline

Baseline thinking

Example 1

Example 2

Example 3

Related Problems

Core Insight

What unlocks the optimal approach

Algorithm Walkthrough

Iteration Checklist

Edge Cases

Full Annotated Code

Interactive Study Demo

Complexity Analysis

Approach Breakdown

Common Mistakes

Off-by-one on range boundaries

Overflow in intermediate arithmetic