# Maximum number of people that can be killed with strength P

There are infinite people standing in a row, indexed from 1. A person having index i has strength of i2. You have strength P and the task is to tell what is the maximum number of people you can kill with strength P.
You can only kill a person with strength X if P ≥ X and after killing him, your strength decreases by X.

Examples:

Input: P = 14
Output: 3
First person will have strength 12 = 1 which is < P
P gets reduced to 13 after the first kill.
Second kill, P = 13 – 22 = 9
Third kill, P = 9 – 32 = 0

Input: P = 58
Output: 5

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive approach: Check every single kill starting from 1 until the strength P is greater than or equal to the strength of the person being killed.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the maximum ` `// number of people that can be killed ` `int` `maxPeople(``int` `p) ` `{ ` `    ``int` `tmp = 0, count = 0; ` ` `  `    ``// Loop will break when the ith person ` `    ``// cannot be killed ` `    ``for` `(``int` `i = 1; i * i <= p; i++) { ` `        ``tmp = tmp + (i * i); ` `        ``if` `(tmp <= p) ` `            ``count++; ` `        ``else` `            ``break``; ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `p = 14; ` `    ``cout << maxPeople(p); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{     ` `// Function to return the maximum ` `// number of people that can be killed ` `static` `int` `maxPeople(``int` `p) ` `{ ` `    ``int` `tmp = ``0``, count = ``0``; ` ` `  `    ``// Loop will break when the ith person ` `    ``// cannot be killed ` `    ``for` `(``int` `i = ``1``; i * i <= p; i++)  ` `    ``{ ` `        ``tmp = tmp + (i * i); ` `        ``if` `(tmp <= p) ` `            ``count++; ` `        ``else` `            ``break``; ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``int` `p = ``14``; ` `    ``System.out.println(maxPeople(p)); ` ` `  `} ` `} ` ` `  `// This code is contributed by Arnab Kundu `

## Python3

 `# Python3 implementation of the approach  ` ` `  `from` `math ``import` `sqrt ` ` `  `# Function to return the maximum  ` `# number of people that can be killed  ` `def` `maxPeople(p) :  ` `     `  `    ``tmp ``=` `0``; count ``=` `0``;  ` ` `  `    ``# Loop will break when the ith person  ` `    ``# cannot be killed  ` `    ``for` `i ``in` `range``(``1``, ``int``(sqrt(p)) ``+` `1``) : ` `        ``tmp ``=` `tmp ``+` `(i ``*` `i);  ` `        ``if` `(tmp <``=` `p) : ` `            ``count ``+``=` `1``;  ` `        ``else` `: ` `            ``break``;  ` `     `  `    ``return` `count;  ` ` `  ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``p ``=` `14``;  ` `    ``print``(maxPeople(p));  ` `     `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{  ` `     `  `// Function to return the maximum ` `// number of people that can be killed ` `static` `int` `maxPeople(``int` `p) ` `{ ` `    ``int` `tmp = 0, count = 0; ` ` `  `    ``// Loop will break when the ith person ` `    ``// cannot be killed ` `    ``for` `(``int` `i = 1; i * i <= p; i++)  ` `    ``{ ` `        ``tmp = tmp + (i * i); ` `        ``if` `(tmp <= p) ` `            ``count++; ` `        ``else` `            ``break``; ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `p = 14; ` `    ``Console.WriteLine(maxPeople(p)); ` `} ` `} ` ` `  `// This code is contributed by anuj_67.. `

Output:

```3
```

Time Complexity: O(N)

Efficient approach: We can see if we kill ith person then we have already killed (i – 1)th person. This means it is a monotonic function f whose domain is the set of integers. Now we can apply binary search on this monotonic function in which instead of array lookup we are now looking for some x such that f(x) is equal to the target value. Time complexity reduces to O(Log(n)).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `#include ` `using` `namespace` `std; ` `#define ll long long ` ` `  `static` `constexpr ``int` `kN = 1000000; ` ` `  `// Function to return the maximum ` `// number of people that can be killed ` `int` `maxPeople(``int` `p) ` `{ ` `    ``// Storing the sum beforehand so that ` `    ``// it can be used in each query ` `    ``ll sums[kN]; ` `    ``sums = 0; ` `    ``for` `(``int` `i = 1; i < kN; i++) ` `        ``sums[i] = (ll)(i * i) + sums[i - 1]; ` ` `  `    ``// lower_bound returns an iterator pointing to the ` `    ``// first element greater than or equal to your val ` `    ``auto` `it = std::lower_bound(sums, sums + kN, p); ` `    ``if` `(*it > p) { ` ` `  `        ``// Previous value ` `        ``--it; ` `    ``} ` ` `  `    ``// Returns the index in array upto which ` `    ``// killing is possible with strength P ` `    ``return` `(it - sums); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `p = 14; ` `    ``cout << maxPeople(p); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG ` `{ ` ` `  `static` `int` `kN = ``1000000``; ` ` `  `// Function to return the maximum ` `// number of people that can be killed ` `static` `int` `maxPeople(``int` `p) ` `{ ` `    ``// Storing the sum beforehand so that ` `    ``// it can be used in each query ` `    ``long` `[]sums = ``new` `long``[kN]; ` `    ``sums[``0``] = ``0``; ` `    ``for` `(``int` `i = ``1``; i < kN; i++) ` `        ``sums[i] = (``long``)(i * i) + sums[i - ``1``]; ` ` `  `    ``// lower_bound returns an iterator pointing to the ` `    ``// first element greater than or equal to your val ` `    ``int` `it = lower_bound(sums, ``0``, kN, p); ` `    ``if` `(it > p)  ` `    ``{ ` ` `  `        ``// Previous value ` `        ``--it; ` `    ``} ` ` `  `    ``// Returns the index in array upto which ` `    ``// killing is possible with strength P ` `    ``return` `it; ` `} ` `private` `static` `int` `lower_bound(``long``[] a, ``int` `low,  ` `                            ``int` `high, ``int` `element) ` `{ ` `    ``while``(low < high) ` `    ``{ ` `        ``int` `middle = low + (high - low)/``2``; ` `        ``if``(element > a[middle]) ` `            ``low = middle + ``1``; ` `        ``else` `            ``high = middle; ` `    ``} ` `    ``return` `low; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `p = ``14``; ` `    ``System.out.println(maxPeople(p)); ` `} ` `} ` ` `  `/* This code is contributed by PrinciRaj1992 */`

## Python3

 `# Python3 implementation of the approach ` `kN ``=` `1000000``; ` ` `  `# Function to return the maximum ` `# number of people that can be killed ` `def` `maxPeople(p): ` ` `  `    ``# Storing the sum beforehand so that ` `    ``# it can be used in each query ` `    ``sums ``=` `[``0``] ``*` `kN; ` `    ``sums[``0``] ``=` `0``; ` `    ``for` `i ``in` `range``(``1``, kN): ` `        ``sums[i] ``=` `(i ``*` `i) ``+` `sums[i ``-` `1``]; ` ` `  `    ``# lower_bound returns an iterator  ` `    ``# pointing to the first element  ` `    ``# greater than or equal to your val ` `    ``it ``=` `lower_bound(sums, ``0``, kN, p); ` `    ``if` `(it > p): ` ` `  `        ``# Previous value ` `        ``it ``-``=` `1``; ` ` `  `    ``# Returns the index in array upto which ` `    ``# killing is possible with strength P ` `    ``return` `it; ` ` `  `def` `lower_bound(a, low, high, element): ` `    ``while``(low < high): ` `        ``middle ``=` `int``(low ``+` `(high ``-` `low) ``/` `2``); ` `        ``if``(element > a[middle]): ` `            ``low ``=` `middle ``+` `1``; ` `        ``else``: ` `            ``high ``=` `middle; ` `    ``return` `low; ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``p ``=` `14``; ` `    ``print``(maxPeople(p)); ` ` `  `# This code contributed by Rajput-Ji `

## C#

 `// C# implementation of the approach ` `using` `System;      ` `     `  `public` `class` `GFG ` `{ ` ` `  `static` `int` `kN = 1000000; ` ` `  `// Function to return the maximum ` `// number of people that can be killed ` `static` `int` `maxPeople(``int` `p) ` `{ ` `    ``// Storing the sum beforehand so that ` `    ``// it can be used in each query ` `    ``long` `[]sums = ``new` `long``[kN]; ` `    ``sums = 0; ` `    ``for` `(``int` `i = 1; i < kN; i++) ` `        ``sums[i] = (``long``)(i * i) + sums[i - 1]; ` ` `  `    ``// lower_bound returns an iterator pointing to the ` `    ``// first element greater than or equal to your val ` `    ``int` `it = lower_bound(sums, 0, kN, p); ` `    ``if` `(it > p)  ` `    ``{ ` ` `  `        ``// Previous value ` `        ``--it; ` `    ``} ` ` `  `    ``// Returns the index in array upto which ` `    ``// killing is possible with strength P ` `    ``return` `it; ` `} ` `private` `static` `int` `lower_bound(``long``[] a, ``int` `low,  ` `                            ``int` `high, ``int` `element) ` `{ ` `    ``while``(low < high) ` `    ``{ ` `        ``int` `middle = low + (high - low)/2; ` `        ``if``(element > a[middle]) ` `            ``low = middle + 1; ` `        ``else` `            ``high = middle; ` `    ``} ` `    ``return` `low; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `p = 14; ` `    ``Console.WriteLine(maxPeople(p)); ` `} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

Output:

```3
```

Time Complexity: O(Log(n))

My Personal Notes arrow_drop_up Competitive Programmer, Full Stack Developer, Technical Content Writer, Machine Learner

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.