Overview

This is an Interval Covering Problem where we need to select the minimum number of given intervals (guard candidates) to cover all \(N\) sections arranged in a line.

Analysis

Key Insight

This problem is a classic greedy algorithm problem: “When completely covering the interval \([1, N]\) on the number line using multiple given intervals, what is the minimum number of intervals required?”

Issues with the Naive Approach

Exhaustively searching all possible selections of guards yields \(2^M\) combinations, and since \(M\) can be up to \(2 \times 10^5\), this is far too slow.

Solution Using Greedy Algorithm

For the interval covering problem, the following greedy strategy is known to yield an optimal solution:

Among the intervals that can cover the leftmost position not yet covered, choose the one with the largest right endpoint.

Let’s consider a concrete example. With \(N = 10\) and intervals \([1,3], [2,6], [4,8], [7,10]\):

1. First, covered = \(0\). We want to cover position \(1\) → Intervals with \(L \leq 1\) are \([1,3]\) and \([2,6]\) (\([2,6]\) has \(L=2 > 1\), so it’s not eligible). Choose \([1,3]\), covered = \(3\).
2. We want to cover position \(4\) → Among intervals with \(L \leq 4\), the one with maximum \(R\) is \([4,8]\) (\(R=8\)) and \([2,6]\) (\(R=6\)). Choose \([4,8]\), covered = \(8\).
3. We want to cover position \(9\) → Among intervals with \(L \leq 9\), choose \([7,10]\) (\(R=10\)), covered = \(10\).

The answer is \(3\) guards.

Algorithm

1. Sort the intervals in ascending order of left endpoint \(L_i\).
2. Prepare a variable covered (the right endpoint of the covered range, initially \(0\)) and a pointer i for scanning the intervals.
3. While covered < N, repeat the following:
   • The next position to cover is covered + 1.
   • Sequentially examine intervals satisfying \(L \leq\) covered + 1, and record the one with the maximum \(R\) as best.
   • If no best is found (or best \(\leq\) covered), covering is impossible, so output \(-1\).
   • Otherwise, set covered = best and increment the number of selected guards by \(1\).
4. After the loop ends, output the number of selected guards.

Key Point: Since we scan the sorted intervals from front to back, the pointer i never goes backward, and each interval is examined at most once overall.

Complexity

• Time complexity: \(O(M \log M)\) (dominated by sorting; the greedy loop itself is \(O(M)\))
• Space complexity: \(O(M)\) (array to store the intervals)

Although \(N\) can be as large as \(10^9\), this is not a problem because the loop depends on the number of intervals \(M\).

Implementation Notes

• covered means “sections \(1\) through covered are covered,” with an initial value of \(0\) (nothing is covered). The next position to cover is covered + 1.

• By using a pointer i after sorting to scan the intervals only once (a two-pointer-like technique), the entire loop is kept to \(O(M)\).

• If coverage cannot be extended (best == -1 or best <= covered), it is determined to be impossible and \(-1\) is returned early.

  Source Code

import sys

def solve():
    input_data = sys.stdin.buffer.read().split()
    idx = 0
    N = int(input_data[idx]); idx += 1
    M = int(input_data[idx]); idx += 1
    
    intervals = []
    for i in range(M):
        L = int(input_data[idx]); idx += 1
        R = int(input_data[idx]); idx += 1
        intervals.append((L, R))
    
    # Greedy interval covering
    # Sort by left endpoint
    intervals.sort()
    
    count = 0
    covered = 0  # we have covered up to 'covered' (0 means nothing covered yet)
    i = 0
    
    while covered < N:
        # We need to cover position covered+1
        # Find the interval with L <= covered+1 that has the maximum R
        best = -1
        while i < M and intervals[i][0] <= covered + 1:
            if intervals[i][1] > best:
                best = intervals[i][1]
            i += 1
        
        if best == -1 or best <= covered:
            # Can't extend coverage
            print(-1)
            return
        
        covered = best
        count += 1
    
    print(count)

solve()

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