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"Solve the following coding problem in C++. I need...

Создано: 25 июня 2026 г.

Отвечено с использованием GPT-5.5 Thinking от Chat01

Вопрос

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"Solve the following coding problem in C++. I need the most optimized solution possible with the lowest time and space complexity. First, analyze the constraints and derive the optimal approach. Avoid brute force and unnecessary data structures. Then provide:

Optimal algorithm explanation (brief).
Time and space complexity.
Clean, concise, competitive-programming style C++17 code.
Use short variable names where readability is not affected.
No redundant comments or extra explanations.
Verify the solution against edge cases and large constraints.

Python

python
from random import randint, seed
seed(0)

def brute(segs,k,N=30):
    # positions 1..N maybe includes all segments
    arr=[0]*(N+2)
    for l,r,c in segs:
        for p in range(l,r+1):
            arr[p]=c
    best=0
    bests=None
    # starts 1..N maybe; include zeros beyond
    for s in range(1,N+1):
        val=sum(arr[s:min(N+1,s+k)])
        if val>best:
            best=val; bests=s
    return best,bests

def calc(a,k):
    a=sorted(a)
    n=len(a)
    best=0
    cur=0
    j=0
    for i in range(n):
        if j < i:
            j=i
            cur=0
        end=a[i][0]+k-1
        while j<n and a[j][1] <= end:
            cur += (a[j][1]-a[j][0]+1)*a[j][2]
            j+=1
        val=cur
        if j<n and a[j][0] <= end:
            val += (end-a[j][0]+1)*a[j][2]
        if val>best: best=val
        if j>i:
            cur -= (a[i][1]-a[i][0]+1)*a[i][2]
        else:
            # current interval not fully in cur; no subtraction
            # j==i
            pass # no need j? let's leave j=i; next i+1 j<i then reset
    return best

def alg(segs,k):
    a=[list(x) for x in segs]
    b=calc(a,k)
    rev=[[-r,-l,c] for l,r,c in segs]
    b=max(b, calc(rev,k))
    return b

# generate non-overlap intervals in 1..20
for n in range(1,7):
  for k in range(1,15):
    for t in range(10000):
      # random subset positions with runs
      pos=1
      segs=[]
      while pos<=20 and len(segs)<n:
        pos += randint(0,3)
        if pos>20: break
        length=randint(1,6)
        r=min(20,pos+length-1)
        c=randint(1,9)
        segs.append((pos,r,c))
        pos=r+randint(1,3)+1
      if not segs: continue
      br,s=brute(segs,k,40)
      al=alg(segs,k)
      if br!=al:
        print("mismatch",n,k,segs,br,s,al)
        raise SystemExit
print("all ok")

Мысль

from random import randint, seed seed(0)

def brute(segs,k,N=30): # positions 1..N maybe includes all segments arr=[0]*(N+2) for l,r,c in segs: for p in range(l,r+1): arr[p]=c best=0 bests=None # starts 1..N maybe; include zeros beyond for s in range(1,N+1): val=sum(arr[s:min(N+1,s+k)]) if val>best: best=val; bests=s return best,bests

def calc(a,k): a=sorted(a) n=len(a) best=0 cur=0 j=0 for i in range(n): if j < i: j=i cur=0 end=a[i][0]+k-1 while j<n and a[j][1] <= end: cur += (a[j][1]-a[j][0]+1)*a[j][2] j+=1 val=cur if j<n and a[j][0] <= end: val += (end-a[j][0]+1)*a[j][2] if val>best: best=val if j>i: cur -= (a[i][1]-a[i][0]+1)*a[i][2] else: # current interval not fully in cur; no subtraction # j==i pass # no need j? let's leave j=i; next i+1 j<i then reset return best

def alg(segs,k): a=[list(x) for x in segs] b=calc(a,k) rev=[[-r,-l,c] for l,r,c in segs] b=max(b, calc(rev,k)) return b

generate non-overlap intervals in 1..20

for n in range(1,7): for k in range(1,15): for t in range(10000): # random subset positions with runs pos=1 segs=[] while pos<=20 and len(segs)<n: pos += randint(0,3) if pos>20: break length=randint(1,6) r=min(20,pos+length-1) c=randint(1,9) segs.append((pos,r,c)) pos=r+randint(1,3)+1 if not segs: continue br,s=brute(segs,k,40) al=alg(segs,k) if br!=al: print("mismatch",n,k,segs,br,s,al) raise SystemExit print("all ok")

Thought for 1m 59s

Ответ

Optimal idea

The money distribution is piecewise constant over non-overlapping segments.
For a window of length k, the maximum can occur when:

  1. The window starts at a segment start, or
  2. The window ends at a segment end

So we sort the segments and use a sliding two-pointer window to compute the best window starting at each segment start.
Then we reverse coordinates and repeat the same logic to cover windows ending at segment ends.

Complexity

  • Sorting: O(n log n)
  • Two scans: O(n)
  • Space: O(n)

Overall: O(n log n) time, O(n) space.

cpp
#include <bits/stdc++.h>
using namespace std;

const long long MOD = 1000000007LL;

long long solve(vector<array<long long,3>> a, long long k) {
    sort(a.begin(), a.end());
    int n = a.size(), j = 0;
    long long cur = 0, ans = 0;

    for (int i = 0; i < n; i++) {
        if (j < i) {
            j = i;
            cur = 0;
        }

        long long end = a[i][0] + k - 1;

        while (j < n && a[j][1] <= end) {
            cur += (a[j][1] - a[j][0] + 1) * a[j][2];
            j++;
        }

        long long val = cur;

        if (j < n && a[j][0] <= end) {
            val += (end - a[j][0] + 1) * a[j][2];
        }

        ans = max(ans, val);

        if (j > i) {
            cur -= (a[i][1] - a[i][0] + 1) * a[i][2];
        }
    }

    return ans;
}

int calculateMaximumConsecutiveSum(int k, vector<vector<int>> segment) {
    vector<array<long long,3>> a;

    for (auto &s : segment) {
        a.push_back({s[0], s[1], s[2]});
    }

    long long ans = solve(a, k);

    vector<array<long long,3>> rev;
    for (auto &x : a) {
        rev.push_back({-x[1], -x[0], x[2]});
    }

    ans = max(ans, solve(rev, k));

    return ans % MOD;
}

Handles:

  • k = 1
  • Very large coordinates up to 1e9
  • Very long segments larger than k
  • Sparse gaps between segments
  • Windows partially covering a segment
  • Maximum answer requiring modulo 1e9 + 7

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