HackerEarth Three rectangles problem solution

YASH PAL, 31 July 202415 February 2026

In this HackerEarth Three rectangles problem solution, You are given a rectangle of height H and width W. You must divide this rectangle exactly into three pieces such that each piece is a rectangle of integral height and width.

You are required to minimize Area(max) – Area(min) where Area(max) is the area of the largest rectangle and Area(min) is the area of the smallest rectangle, among all three rectangle pieces.

HackerEarth Three rectangles problem solution

HackerEarth Three rectangles problem solution.

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

#define     F           first
#define     S           second
#define     pb          push_back
#define     lb          lower_bound
#define     ub          upper_bound
#define     vi          vector<int>
#define     all(x)      x.begin(),x.end()
#define     fix         fixed<<setprecision(10)
#define     rep(i,a,b)  for(int i=int(a);i<=int(b);i++)
#define     repb(i,b,a) for(int i=int(b);i>=int(a);i--)
#define     FastIO      ios_base::sync_with_stdio(0),cin.tie(0)

typedef double db;
typedef long long ll;

const int N=2e5+5;
const int mod=1e9+7;

void solve(){
    ll h,w;
    cin>>h>>w;
    ll ans=1e18;
    rep(i,1,h-1){
        ll nh=h-i,nw=w;
        ll maxi=max({i*nw,nh/2*nw,(nh-nh/2)*nw});
        ll mini=min({i*nw,nh/2*nw,(nh-nh/2)*nw});
        ans=min(ans,maxi-mini);
        maxi=max({i*nw,nw/2*nh,(nw-nw/2)*nh});
        mini=min({i*nw,nw/2*nh,(nw-nw/2)*nh});
        ans=min(ans,maxi-mini);
    }
    rep(i,1,w-1){
        ll nw=w-i,nh=h;
        ll maxi=max({i*nh,nw/2*nh,(nw-nw/2)*nh});
        ll mini=min({i*nh,nw/2*nh,(nw-nw/2)*nh});
        ans=min(ans,maxi-mini);
        maxi=max({i*nh,nh/2*nw,(nh-nh/2)*nw});
        mini=min({i*nh,nh/2*nw,(nh-nh/2)*nw});
        ans=min(ans,maxi-mini);
    }
    cout<<ans<<'n';
}

signed main(){
    FastIO;
    int t;
    cin>>t;
    while(t--) solve();
    return 0;
}

Second solution

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

const int N = 1e5 + 14;

int solve(int h, int w) {
    int ans = INT_MAX;
    for (auto x : {(ll) w * h / (3 * h / 2), (ll) w * h / (3 * h / 2) + 1})
        ans = min<ll>(ans, max({abs(x * ll(h / 2) - ll(w - x) * h), x * ll((h + 1) / 2) - ll(w - x) * h, h % 2 * x}));
    return ans;
}

int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
        int h, w;
        cin >> h >> w;
        cout << min({h % 3 == 0 || w % 3 == 0 ? 0 : min(h, w), solve(h, w), solve(w, h)}) << 'n';
    }
}

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