# 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))

More Efficient Approach :
We can do the same problem in time complexity O(logn) and Space Complexity in O(1). Start your binary search by considering the value of low as 0 and high as 10^15. We will calculate the mid-value and according to mid, we will change the position of low and high.
Below is the implementation of the above approach.

## Python3

 `# Python3 implementation of the approach ` ` `  `# helper function which returns the sum ` `# of series (1^2 + 2^2 +...+ n^2) ` `def` `squareSeries(n): ` `    ``return` `(n``*``(n``+``1``)``*``(``2``*``n``+``1``))``/``/``6` ` `  `# maxPeople function which returns ` `# appropriate value using Binary Search ` `# in O(logn) ` ` `  `def` `maxPeople(n): ` ` `  `    ``# Set the lower and higher values ` `    ``low ``=` `0` `    ``high ``=` `1000000000000000` `    ``while` `low<``=``high: ` ` `  `        ``# calculate the mid using ` `        ``# low and high ` ` `  `        ``mid ``=` `low ``+` `((high``-``low)``/``/``2``) ` `        ``value ``=` `squareSeries(mid) ` ` `  `        ``#compare value with n ` `        ``if` `value<``=``n: ` `            ``ans ``=` `mid ` `            ``low ``=` `mid``+``1` `        ``else``: ` `            ``high ``=` `mid``-``1` ` `  `    ``# return the ans ` `    ``return` `ans ` ` `  `if` `__name__``=``=``'__main__'``: ` `    ``p``=``14` `    ``print``(maxPeople(p)) ` ` `  `# This code is contributed bu chaudhary_19 ` `# (* Mayank Chaudhary) `

Output:

```3
```

Time Complexity: O(Log(n))
Space Complexity: O(1)

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