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C++ Program to check cats words are right or not with colored hats
Suppose we have an array A with N elements. Consider there are N cats and they are numbered from 1 to N. Each cat is wearing a hat and ith cat says "there are exactly A[i] number of different colors among the N-1 hats owned by the cats except me". We have to check whether there exists a sequence of colors of the hats which is consistent with the remarks of the cats.
So, if the input is like A = [1, 2, 2], then the output will be True, because if cat 1, 2 and 3 wears hats in colors say red, blue and blue hats, respectively, it is consistent with the remarks of the cats.
To solve this, we will follow these steps −
mn := inf, mx = 0, cnt = 0
n := size of A
Define an array a of size (n + 1)
for initialize i := 1, when i <= n, update (increase i by 1), do:
a[i] := A[i - 1]
mn := minimum of mn and a[i]
mx = maximum of mx and a[i]
for initialize i := 1, when i <= n, update (increase i by 1), do:
if a[i] is same as mn, then:
(increase cnt by 1)
if mx is same as mn, then:
if mn is same as n - 1 or 2 * mn <= n, then:
return true
Otherwise
return false
otherwise when mx is same as mn + 1, then:
if mn >= cnt and n - cnt >= 2 * (mx - cnt), then:
return true
Otherwise
return false
Otherwise
return falseExample
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
bool solve(vector<int> A) {
int mn = 99999, mx = 0, cnt = 0;
int n = A.size();
vector<int> a(n + 1);
for (int i = 1; i <= n; ++i) {
a[i] = A[i - 1];
mn = min(mn, a[i]), mx = max(mx, a[i]);
}
for (int i = 1; i <= n; ++i)
if (a[i] == mn)
++cnt;
if (mx == mn) {
if (mn == n - 1 || 2 * mn <= n)
return true;
else
return false;
}
else if (mx == mn + 1) {
if (mn >= cnt && n - cnt >= 2 * (mx - cnt))
return true;
else
return false;
}
else
return false;
}
int main() {
vector<int> A = { 1, 2, 2 };
cout << solve(A) << endl;
}Input
{ 1, 2, 2 }Output
1
Updated on: 25-Feb-2022
115 Views
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