LeetCode #1562 — MEDIUM

Find Latest Group of Size M

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

The Problem

Problem Statement

Given an array arr that represents a permutation of numbers from 1 to n.

You have a binary string of size n that initially has all its bits set to zero. At each step i (assuming both the binary string and arr are 1-indexed) from 1 to n, the bit at position arr[i] is set to 1.

You are also given an integer m. Find the latest step at which there exists a group of ones of length m. A group of ones is a contiguous substring of 1's such that it cannot be extended in either direction.

Return the latest step at which there exists a group of ones of length exactly m. If no such group exists, return -1.

Example 1:

Input: arr = [3,5,1,2,4], m = 1
Output: 4
Explanation: 
Step 1: "00100", groups: ["1"]
Step 2: "00101", groups: ["1", "1"]
Step 3: "10101", groups: ["1", "1", "1"]
Step 4: "11101", groups: ["111", "1"]
Step 5: "11111", groups: ["11111"]
The latest step at which there exists a group of size 1 is step 4.

Example 2:

Input: arr = [3,1,5,4,2], m = 2
Output: -1
Explanation: 
Step 1: "00100", groups: ["1"]
Step 2: "10100", groups: ["1", "1"]
Step 3: "10101", groups: ["1", "1", "1"]
Step 4: "10111", groups: ["1", "111"]
Step 5: "11111", groups: ["11111"]
No group of size 2 exists during any step.

Constraints:

n == arr.length
1 <= m <= n <= 10⁵
1 <= arr[i] <= n
All integers in arr are distinct.

Step 01

Brute Force Baseline

Problem summary: Given an array arr that represents a permutation of numbers from 1 to n. You have a binary string of size n that initially has all its bits set to zero. At each step i (assuming both the binary string and arr are 1-indexed) from 1 to n, the bit at position arr[i] is set to 1. You are also given an integer m. Find the latest step at which there exists a group of ones of length m. A group of ones is a contiguous substring of 1's such that it cannot be extended in either direction. Return the latest step at which there exists a group of ones of length exactly m. If no such group exists, return -1.

Baseline thinking

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

Pattern signal: Array · Hash Map · Binary Search

Example 1

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

Example 2

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

Step 02

Core Insight

What unlocks the optimal approach

Since the problem asks for the latest step, can you start the searching from the end of arr?
Use a map to store the current “1” groups.
At each step (going backwards) you need to split one group and update the map.

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 #1562: Find Latest Group of Size M
class Solution {
    private int[] p;
    private int[] size;

    public int findLatestStep(int[] arr, int m) {
        int n = arr.length;
        if (m == n) {
            return n;
        }
        boolean[] vis = new boolean[n];
        p = new int[n];
        size = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
            size[i] = 1;
        }
        int ans = -1;
        for (int i = 0; i < n; ++i) {
            int v = arr[i] - 1;
            if (v > 0 && vis[v - 1]) {
                if (size[find(v - 1)] == m) {
                    ans = i;
                }
                union(v, v - 1);
            }
            if (v < n - 1 && vis[v + 1]) {
                if (size[find(v + 1)] == m) {
                    ans = i;
                }
                union(v, v + 1);
            }
            vis[v] = true;
        }
        return ans;
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }

    private void union(int a, int b) {
        int pa = find(a), pb = find(b);
        if (pa == pb) {
            return;
        }
        p[pa] = pb;
        size[pb] += size[pa];
    }
}

// Accepted solution for LeetCode #1562: Find Latest Group of Size M
func findLatestStep(arr []int, m int) int {
	n := len(arr)
	if m == n {
		return n
	}
	p := make([]int, n)
	size := make([]int, n)
	vis := make([]bool, n)
	for i := range p {
		p[i] = i
		size[i] = 1
	}
	var find func(int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}
	union := func(a, b int) {
		pa, pb := find(a), find(b)
		if pa == pb {
			return
		}
		p[pa] = pb
		size[pb] += size[pa]
	}

	ans := -1
	for i, v := range arr {
		v--
		if v > 0 && vis[v-1] {
			if size[find(v-1)] == m {
				ans = i
			}
			union(v, v-1)
		}
		if v < n-1 && vis[v+1] {
			if size[find(v+1)] == m {
				ans = i
			}
			union(v, v+1)
		}
		vis[v] = true
	}
	return ans
}

# Accepted solution for LeetCode #1562: Find Latest Group of Size M
class Solution:
    def findLatestStep(self, arr: List[int], m: int) -> int:
        def find(x):
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        def union(a, b):
            pa, pb = find(a), find(b)
            if pa == pb:
                return
            p[pa] = pb
            size[pb] += size[pa]

        n = len(arr)
        if m == n:
            return n
        vis = [False] * n
        p = list(range(n))
        size = [1] * n
        ans = -1
        for i, v in enumerate(arr):
            v -= 1
            if v and vis[v - 1]:
                if size[find(v - 1)] == m:
                    ans = i
                union(v, v - 1)
            if v < n - 1 and vis[v + 1]:
                if size[find(v + 1)] == m:
                    ans = i
                union(v, v + 1)
            vis[v] = True
        return ans

// Accepted solution for LeetCode #1562: Find Latest Group of Size M
struct Solution;

use std::collections::HashMap;

struct UnionFind {
    parent: Vec<usize>,
    size: Vec<usize>,
    group: HashMap<usize, usize>,
    n: usize,
}

impl UnionFind {
    fn new(n: usize) -> Self {
        let parent = (0..n).collect();
        let size = vec![0; n];
        let group = HashMap::new();
        UnionFind {
            parent,
            size,
            group,
            n,
        }
    }

    fn set_one(&mut self, i: usize) {
        self.size[i] = 1;
        *self.group.entry(1).or_default() += 1;
    }

    fn find(&mut self, i: usize) -> usize {
        let j = self.parent[i];
        if i == j {
            i
        } else {
            let k = self.find(j);
            self.parent[i] = k;
            k
        }
    }

    fn union(&mut self, mut i: usize, mut j: usize) {
        i = self.find(i);
        j = self.find(j);
        let a = self.size[i];
        let b = self.size[j];
        let c = a + b;
        *self.group.entry(a).or_default() -= 1;
        *self.group.entry(b).or_default() -= 1;
        *self.group.entry(c).or_default() += 1;
        if i != j {
            self.parent[i] = j;
            self.size[j] += self.size[i];
            self.n -= 1;
        }
    }
}

impl Solution {
    fn find_latest_step(arr: Vec<i32>, m: i32) -> i32 {
        let m = m as usize;
        let n = arr.len();
        let mut values = vec![0; n];
        let mut res = -1;
        let mut uf = UnionFind::new(n);
        for i in 0..n {
            let j = (arr[i] - 1) as usize;
            values[j] = 1;
            uf.set_one(j);
            if j > 0 && values[j - 1] == 1 {
                uf.union(j - 1, j);
            }
            if j + 1 < n && values[j + 1] == 1 {
                uf.union(j, j + 1);
            }
            if *uf.group.entry(m).or_default() > 0 {
                res = (i + 1) as i32;
            }
        }
        res
    }
}

#[test]
fn test() {
    let arr = vec![3, 5, 1, 2, 4];
    let m = 1;
    let res = 4;
    assert_eq!(Solution::find_latest_step(arr, m), res);
}

// Accepted solution for LeetCode #1562: Find Latest Group of Size M
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1562: Find Latest Group of Size M
// class Solution {
//     private int[] p;
//     private int[] size;
// 
//     public int findLatestStep(int[] arr, int m) {
//         int n = arr.length;
//         if (m == n) {
//             return n;
//         }
//         boolean[] vis = new boolean[n];
//         p = new int[n];
//         size = new int[n];
//         for (int i = 0; i < n; ++i) {
//             p[i] = i;
//             size[i] = 1;
//         }
//         int ans = -1;
//         for (int i = 0; i < n; ++i) {
//             int v = arr[i] - 1;
//             if (v > 0 && vis[v - 1]) {
//                 if (size[find(v - 1)] == m) {
//                     ans = i;
//                 }
//                 union(v, v - 1);
//             }
//             if (v < n - 1 && vis[v + 1]) {
//                 if (size[find(v + 1)] == m) {
//                     ans = i;
//                 }
//                 union(v, v + 1);
//             }
//             vis[v] = true;
//         }
//         return ans;
//     }
// 
//     private int find(int x) {
//         if (p[x] != x) {
//             p[x] = find(p[x]);
//         }
//         return p[x];
//     }
// 
//     private void union(int a, int b) {
//         int pa = find(a), pb = find(b);
//         if (pa == pb) {
//             return;
//         }
//         p[pa] = pb;
//         size[pb] += size[pa];
//     }
// }

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(log n)

Space

O(1)

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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.