Find the winner of a game of removing at most 3 stones from a pile in each turn
Last Updated :
23 Jul, 2025
Given an array arr[] of size N, denoting values assigned to N stones, two players, Player1 and Player2, play a game of alternating turns. In each turn, a player can take 1, 2, or 3 stones from the first remaining stones with the sum of values of all the removed stones added to the player's score. Considering that both players play optimally, the task is to print the winner of the game. If both players end the game with the same score, print "Tie".
Examples:
Input: arr[] = {1, 2, 3, 7}
Output: Player2
Explanation: Player1 will always lose in an optimal scenario.
Input: arr[] = {1, 2, 3, -9}
Output: Player1
Explanation: Player1 must choose all the three piles at the first move to win and leave Player2 with negative score.
If Player1 chooses only one stone his score will be 1 and the next move Player2 score becomes 5.
The next move Player1 will take the stone with value = -9 and lose.
If Player1 chooses two piles his score will be 3 and the next move Player2 score becomes 3.
The next move Player1 will take the stone with value = -9 and also lose.
Naive Approach: The simple approach is to pick the number of stones that will maximize the total sum of the stone's values. As both players play optimally and Player1 starts the game, Player1 picks either 1 or 2 or 3 stones and the remaining stones passed to the next player. Therefore, the score of Player2 must be subtracted from score of Player1. The idea is to use recursion to solve the problem. Let the maximum score of Player1 be res which is obtained after the recursive calls.
- If the result, res > 0, Player1 wins
- If the result, res < 0, Player2 wins
- If the result, res == 0, then it is a tie.
The recursive tree looks like this, where some subproblems are repeated many times.
Follow the steps to solve the problem:
- Declare a recursive function, say maxScore(i), to calculate the maximum score of Player1 if the game starts at index i
- If the value of i ≥ n, return 0.
- Initialize a variable, say score as INT_MIN, to store the maximum score of Player1
- Picks 1 stone: score = max(score, arr[i] - maxScore(i + 1))
- Picks 2 stones i.e (i + 1 < N): score = max(score, arr[i] + arr[i + 1] - maxScore(i + 2))
- Picks 3 stones i.e (i + 2 < N): score = max(score, arr[i] + arr[i + 1] + arr[i + 2] - maxScore(i + 3))
- Return the value of the score.
- Store the value of maxScore(0) in a variable res.
- If the value of res>0, print "Player1", if res<0, print "Player2", otherwise print "Tie".
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the maximum score of Player1
int maximumStonesUtil(int* arr, int n, int i)
{
// Base Case
if (i >= n)
return 0;
// Variable to store maximum score
int ans = INT_MIN;
// Pick one stone
ans = max(ans,
arr[i] - maximumStonesUtil(arr, n, i + 1));
// Pick 2 stones
if (i + 1 < n)
ans = max(ans,
arr[i] + arr[i + 1]
- maximumStonesUtil(arr, n, i + 2));
// Pick 3 stones
if (i + 2 < n)
ans = max(ans,
arr[i] + arr[i + 1] + arr[i + 2]
- maximumStonesUtil(arr, n, i + 3));
// Return the score of the player
return ans;
}
// Function to find the winner of the game
string maximumStones(int* arr, int n)
{
// Store the result
int res = maximumStonesUtil(arr, n, 0);
// Player 1 wins
if (res > 0)
return "Player1";
// PLayer 2 wins
else if (res < 0)
return "Player2";
// Tie
else
return "Tie";
}
// Driver Code
int main()
{
// Given Input
int arr[] = { 1, 2, 3, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function Call
cout << maximumStones(arr, n);
return 0;
}
// Java program for the above approach
class GFG{
// Function to find the maximum score of Player1
static int maximumStonesUtil(int[] arr, int n,
int i)
{
// Base Case
if (i >= n)
return 0;
// Variable to store maximum score
int ans = Integer.MIN_VALUE;
// Pick one stone
ans = Math.max(ans, arr[i] - maximumStonesUtil(
arr, n, i + 1));
// Pick 2 stones
if (i + 1 < n)
ans = Math.max(ans, arr[i] + arr[i + 1] -
maximumStonesUtil(arr, n, i + 2));
// Pick 3 stones
if (i + 2 < n)
ans = Math.max(
ans,
arr[i] + arr[i + 1] + arr[i + 2]
- maximumStonesUtil(arr, n, i + 3));
// Return the score of the player
return ans;
}
// Function to find the winner of the game
static String maximumStones(int[] arr, int n)
{
// Store the result
int res = maximumStonesUtil(arr, n, 0);
// Player 1 wins
if (res > 0)
return "Player1";
// PLayer 2 wins
else if (res < 0)
return "Player2";
// Tie
else
return "Tie";
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 2, 3, 7 };
int n = arr.length;
// Function Call
System.out.println(maximumStones(arr, n));
}
}
// This code is contributed by abhinavjain194
# Python3 program for the above approach
import sys
# Function to find the maximum score of Player1
def maximumStonesUtil(arr, n, i):
# Base Case
if (i >= n):
return 0
# Variable to store maximum score
ans = -sys.maxsize-1;
# Pick one stone
ans = max(
ans, arr[i] - maximumStonesUtil(
arr, n, i + 1))
# Pick 2 stones
if (i + 1 < n):
ans = max(
ans, arr[i] + arr[i + 1]- maximumStonesUtil(
arr, n, i + 2))
# Pick 3 stones
if (i + 2 < n):
ans = max(
ans, arr[i] + arr[i + 1] + arr[i + 2]-
maximumStonesUtil(arr, n, i + 3));
# Return the score of the player
return ans
# Function to find the winner of the game
def maximumStones(arr, n):
# Store the result
res = maximumStonesUtil(arr, n, 0)
# Player 1 wins
if (res > 0):
return "Player1"
# PLayer 2 wins
elif(res < 0):
return "Player2"
# Tie
else:
return "Tie"
# Driver Code
if __name__ == '__main__':
# Given Input
arr = [ 1, 2, 3, 7 ]
n = len(arr)
# Function Call
print(maximumStones(arr, n))
# This code is contributed by SURENDRA_GANGWAR
// C# program for the above approach
using System;
class GFG
{
// Function to find the maximum score of Player1
static int maximumStonesUtil(int[] arr, int n,
int i)
{
// Base Case
if (i >= n)
return 0;
// Variable to store maximum score
int ans = Int32.MinValue;
// Pick one stone
ans = Math.Max(ans, arr[i] - maximumStonesUtil(
arr, n, i + 1));
// Pick 2 stones
if (i + 1 < n)
ans = Math.Max(ans, arr[i] + arr[i + 1] -
maximumStonesUtil(arr, n, i + 2));
// Pick 3 stones
if (i + 2 < n)
ans = Math.Max(
ans,
arr[i] + arr[i + 1] + arr[i + 2]
- maximumStonesUtil(arr, n, i + 3));
// Return the score of the player
return ans;
}
// Function to find the winner of the game
static String maximumStones(int[] arr, int n)
{
// Store the result
int res = maximumStonesUtil(arr, n, 0);
// Player 1 wins
if (res > 0)
return "Player1";
// PLayer 2 wins
else if (res < 0)
return "Player2";
// Tie
else
return "Tie";
}
// Driver Code
public static void Main()
{
int[] arr = { 1, 2, 3, 7 };
int n = arr.Length;
// Function Call
Console.WriteLine(maximumStones(arr, n));
}
}
// This code is contributed by code_hunt.
<script>
// Javascript program for the above approach
// Function to find the maximum score of Player1
function maximumStonesUtil(arr, n, i)
{
// Base Case
if (i >= n)
return 0;
// Variable to store maximum score
let ans = Number.MIN_VALUE;
// Pick one stone
ans = Math.max(ans, arr[i] - maximumStonesUtil(arr, n, i + 1));
// Pick 2 stones
if (i + 1 < n)
ans = Math.max(ans, arr[i] + arr[i + 1] - maximumStonesUtil(arr, n, i + 2));
// Pick 3 stones
if (i + 2 < n)
ans = Math.max(
ans,
arr[i] + arr[i + 1] + arr[i + 2]
- maximumStonesUtil(arr, n, i + 3));
// Return the score of the player
return ans;
}
// Function to find the winner of the game
function maximumStones(arr, n)
{
// Store the result
let res = maximumStonesUtil(arr, n, 0);
// Player 1 wins
if (res < 0)
return "Player1";
// PLayer 2 wins
else if (res > 0)
return "Player2";
// Tie
else
return "Tie";
}
let arr = [ 1, 2, 3, 7 ];
let n = arr.length;
// Function Call
document.write(maximumStones(arr, n));
// This code is contributed by rameshtravel07.
</script>
Time Complexity: O(3N)
Auxiliary Space: O(N)
Efficient Approach: To optimize the above approach, the idea is to use dynamic programming. The given problem has optimal substructure property and overlapping subproblems. Use memoization by creating a 1D table, dp of size N to store the results of the recursive calls.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the maximum score of Player 1
int maximumStonesUtil(int* arr, int n, int i,
vector<int>& dp)
{
// Base Case
if (i >= n)
return 0;
int& ans = dp[i];
// If the result is already computed, then
// return the result
if (ans != -1)
return ans;
// Variable to store maximum score
ans = INT_MIN;
// Pick one stone
ans = max(
ans, arr[i] - maximumStonesUtil(arr, n, i + 1, dp));
// Pick 2 stones
if (i + 1 < n)
ans = max(ans, arr[i] + arr[i + 1]
- maximumStonesUtil(arr, n,
i + 2, dp));
// Pick 3 stones
if (i + 2 < n)
ans = max(ans, arr[i] + arr[i + 1] + arr[i + 2]
- maximumStonesUtil(arr, n,
i + 3, dp));
// Return the score of the player
return ans;
}
// Function to find the winner of the game
string maximumStones(int* arr, int n)
{
// Create a 1D table, dp of size N
vector<int> dp(n, -1);
// Store the result
int res = maximumStonesUtil(arr, n, 0, dp);
// Player 1 wins
if (res > 0)
return "Player1";
// PLayer 2 wins
else if (res < 0)
return "Player2";
// Tie
else
return "Tie";
}
// Driver Code
int main()
{
// Given Input
int arr[] = { 1, 2, 3, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function Call
cout << maximumStones(arr, n);
return 0;
}
// Java program for the above approach
import java.util.*;
class GFG
{
static int ans = 0;
// Function to find the maximum score of Player 1
static int maximumStonesUtil(int[] arr, int n, int i,
int[] dp)
{
// Base Case
if (i >= n)
return 0;
ans = dp[i];
// If the result is already computed, then
// return the result
if (ans != -1)
return ans;
// Variable to store maximum score
ans = Integer.MIN_VALUE;
// Pick one stone
ans = Math.max(
ans, arr[i] - maximumStonesUtil(arr, n, i + 1, dp));
// Pick 2 stones
if (i + 1 < n)
ans = Math.max(ans, arr[i] + arr[i + 1]
- maximumStonesUtil(arr, n,
i + 2, dp));
// Pick 3 stones
if (i + 2 < n)
ans = Math.max(ans, arr[i] + arr[i + 1] + arr[i + 2]
- maximumStonesUtil(arr, n,
i + 3, dp));
// Return the score of the player
return ans;
}
// Function to find the winner of the game
static String maximumStones(int []arr, int n)
{
// Create a 1D table, dp of size N
int []dp = new int[n];
Arrays.fill(dp, -1);
// Store the result
int res = maximumStonesUtil(arr, n, 0, dp);
// Player 1 wins
if (res > 0)
return "Player1";
// PLayer 2 wins
else if (res < 0)
return "Player2";
// Tie
else
return "Tie";
}
// Driver Code
public static void main(String[] args)
{
// Given Input
int arr[] = { 1, 2, 3, 7 };
int n = arr.length;
// Function Call
System.out.print(maximumStones(arr, n));
}
}
// This code is contributed by Amit Katiyar
# Python program for the above approach
# Function to find the maximum score of Player 1
def maximumStonesUtil(arr, n, i, dp):
# Base Case
if (i >= n):
return 0
ans = dp[i]
# If the result is already computed, then
# return the result
if (ans != -1):
return ans
# Variable to store maximum score
ans = -2**31
# Pick one stone
ans = max( ans, arr[i] - maximumStonesUtil(arr, n, i + 1, dp))
# Pick 2 stones
if (i + 1 < n):
ans = max(ans, arr[i] + arr[i + 1] - maximumStonesUtil(arr, n, i + 2, dp))
# Pick 3 stones
if (i + 2 < n):
ans = max(ans, arr[i] + arr[i + 1] + arr[i + 2] - maximumStonesUtil(arr, n, i + 3, dp))
# Return the score of the player
return ans
# Function to find the winner of the game
def maximumStones(arr, n):
# Create a 1D table, dp of size N
dp =[-1]*n
# Store the result
res = maximumStonesUtil(arr, n, 0, dp)
# Player 1 wins
if (res > 0):
return "Player1"
# PLayer 2 wins
elif (res < 0):
return "Player2"
# Tie
else:
return "Tie"
# Driver Code
# Given Input
arr = [1, 2, 3, 7]
n = len(arr)
# Function Call
print(maximumStones(arr, n))
# This code is contributed by shivani
// C# program for the above approach
using System;
public class GFG
{
static int ans = 0;
// Function to find the maximum score of Player 1
static int maximumStonesUtil(int[] arr, int n, int i,
int[] dp)
{
// Base Case
if (i >= n)
return 0;
ans = dp[i];
// If the result is already computed, then
// return the result
if (ans != -1)
return ans;
// Variable to store maximum score
ans = int.MinValue;
// Pick one stone
ans = Math.Max(
ans, arr[i] - maximumStonesUtil(arr, n, i + 1, dp));
// Pick 2 stones
if (i + 1 < n)
ans = Math.Max(ans, arr[i] + arr[i + 1]
- maximumStonesUtil(arr, n,
i + 2, dp));
// Pick 3 stones
if (i + 2 < n)
ans = Math.Max(ans, arr[i] + arr[i + 1] + arr[i + 2]
- maximumStonesUtil(arr, n,
i + 3, dp));
// Return the score of the player
return ans;
}
// Function to find the winner of the game
static String maximumStones(int []arr, int n)
{
// Create a 1D table, dp of size N
int []dp = new int[n];
for(int i = 0; i < n; i++)
dp[i]=-1;
// Store the result
int res = maximumStonesUtil(arr, n, 0, dp);
// Player 1 wins
if (res > 0)
return "Player1";
// PLayer 2 wins
else if (res < 0)
return "Player2";
// Tie
else
return "Tie";
}
// Driver Code
public static void Main(String[] args)
{
// Given Input
int []arr = { 1, 2, 3, 7 };
int n = arr.Length;
// Function Call
Console.Write(maximumStones(arr, n));
}
}
// This code contributed by shikhasingrajput
<script>
// Javascript program for the above approach
let ans = 0 ;
// Function to find the maximum score of Player 1
function maximumStonesUtil( arr, n, i,
dp)
{
// Base Case
if (i >= n)
return 0;
ans = dp[i];
// If the result is already computed, then
// return the result
if (ans != -1)
return ans;
// Variable to store maximum score
ans = -Math.pow(2,31);
// Pick one stone
ans = Math.max(
ans, arr[i] - maximumStonesUtil(arr, n, i + 1, dp));
// Pick 2 stones
if (i + 1 < n)
ans = Math.max(ans, arr[i] + arr[i + 1]
- maximumStonesUtil(arr, n,
i + 2, dp));
// Pick 3 stones
if (i + 2 < n)
ans = Math.max(ans, arr[i] + arr[i + 1] + arr[i + 2]
- maximumStonesUtil(arr, n,
i + 3, dp));
// Return the score of the player
return ans;
}
// Function to find the winner of the game
function maximumStones(arr, n)
{
// Create a 1D table, dp of size N
let dp = new Array(n).fill(-1);
// Store the result
let res = maximumStonesUtil(arr, n, 0, dp);
// Player 1 wins
if (res > 0)
return "Player1";
// PLayer 2 wins
else if (res < 0)
return "Player2";
// Tie
else
return "Tie";
}
// Driver Code
let arr= [ 1, 2, 3, 7 ];
let n = arr.length;
// Function Call
document.write(maximumStones(arr, n));
// This code is contributed by jana_sayantan.
<script>
Time Complexity: O(N)
Auxiliary Space: O(N)
Efficient Approach: Using the DP Tabulation method ( Iterative approach )
The approach to solving this problem is the same but DP tabulation(bottom-up) method is better than the Dp + memoization(top-down) because the memoization method needs extra stack space of recursion calls.
Steps to solve this problem:
- Create a DP of size n+3 to store the solution of the subproblems.
- Initialize the DP with base cases dp[n] = dp[n+1] = dp[n+2] = 0.
- Now Iterate over subproblems to get the value of the current problem from the previous computation of subproblems stored in DP
- Print the final result
- if dp[0] > 0 print player1.
- if dp[0] < 0 print player2.
- else print tie.
Implementation:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the winner of game
string maximumStones(int* arr, int n)
{
// Create a 1D table, dp of size N+3
vector<int> dp(n + 3);
// Initialize the table for the
// last 3 stones
dp[n] = dp[n + 1] = dp[n + 2] = 0;
// Calculate the table for the
// remaining stones in reverse order
for (int i = n - 1; i >= 0; i--) {
// Pick one stone
int a = arr[i] - dp[i + 1];
int b = -1 , c= -1;
if(i+1 == n){
b = arr[i]- dp[i + 2];
c = arr[i] - dp[i + 3];
}
else if ( i+2 == n){
c = arr[i] + arr[i + 1] - dp[i + 3];
}
if(b==-1)
b = arr[i] + arr[i + 1] - dp[i + 2];
if(c == -1)
c = arr[i] + arr[i + 1] + arr[i + 2] - dp[i + 3];
dp[i] = max(a,max(b,c));
}
// Player 1 wins
if (dp[0] > 0)
return "Player1";
// PLayer 2 wins
else if (dp[0] < 0)
return "Player2";
// Tie
else
return "Tie";
}
// Driver Code
int main()
{
// Given Input
int arr[] = { 1, 2, 3, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function Call
cout << maximumStones(arr, n);
return 0;
}
import java.util.*;
public class Main {
// Function to find the winner of the game
public static String maximumStones(int[] arr, int n) {
// Create a 1D table, dp of size N+3
int[] dp = new int[n + 3];
// Initialize the table for the
// last 3 stones
dp[n] = dp[n + 1] = dp[n + 2] = 0;
// Calculate the table for the
// remaining stones in reverse order
for (int i = n - 1; i >= 0; i--) {
// Pick one stone
int a = arr[i] - dp[i + 1];
int b = -1, c = -1;
if (i + 1 == n) {
b = arr[i] - dp[i + 2];
c = arr[i] - dp[i + 3];
} else if (i + 2 == n) {
c = arr[i] + arr[i + 1] - dp[i + 3];
}
if (b == -1)
b = arr[i] + arr[i + 1] - dp[i + 2];
if (c == -1)
c = arr[i] + arr[i + 1] + arr[i + 2] - dp[i + 3];
dp[i] = Math.max(a, Math.max(b, c));
}
// Player 1 wins
if (dp[0] > 0)
return "Player1";
// PLayer 2 wins
else if (dp[0] < 0)
return "Player2";
// Tie
else
return "Tie";
}
// Driver Code
public static void main(String[] args) {
// Given Input
int[] arr = { 1, 2, 3, 7 };
int n = arr.length;
// Function Call
System.out.println(maximumStones(arr, n));
}
}
# Function to find the winner of the game
def maximumStones(arr, n):
# Create a 1D table, dp of size N+3
dp = [0] * (n+3)
# Initialize the table for the
# last 3 stones
dp[n] = dp[n + 1] = dp[n + 2] = 0
# Calculate the table for the
# remaining stones in reverse order
for i in range(n-1, -1, -1):
# Pick one stone
a = arr[i] - dp[i + 1]
b = c = -1
if i+1 == n:
b = arr[i] - dp[i + 2]
c = arr[i] - dp[i + 3]
elif i+2 == n:
c = arr[i] + arr[i + 1] - dp[i + 3]
if b == -1:
b = arr[i] + arr[i + 1] - dp[i + 2]
if c == -1:
c = arr[i] + arr[i + 1] + arr[i + 2] - dp[i + 3]
dp[i] = max(a, max(b, c))
# Player 1 wins
if dp[0] > 0:
return "Player1"
# Player 2 wins
elif dp[0] < 0:
return "Player2"
# Tie
else:
return "Tie"
# Driver Code
if __name__ == '__main__':
# Given Input
arr = [1, 2, 3, 7]
n = len(arr)
# Function Call
print(maximumStones(arr, n))
using System;
class GFG {
// Function to find the winner of the game "Maximum Stones"
// and return "Player1", "Player2", or "Tie"
public static string MaximumStones(int[] arr, int n) {
// Create an array to store the maximum stones each player
// can collect starting from the i-th position
int[] dp = new int[n + 3];
// Initialize the last three elements as
// 0 (no stones available after the last position)
dp[n] = dp[n + 1] = dp[n + 2] = 0;
// Iterate through the array from right to left to
// calculate the maximum stones each player can collect
for (int i = n - 1; i >= 0; i--) {
// Calculate the maximum stones Player1 can collect
// starting from the i-th position
int a = arr[i] - dp[i + 1];
// Initialize variables to store the maximum stones
// Player2 can collect from the next two positions
int b = -1, c = -1;
// If there is only one position left after the i-th
// position, calculate the maximum stones Player2 can collect
// from that position
if (i + 1 == n) {
b = arr[i] - dp[i + 2];
c = arr[i] - dp[i + 3];
}
// If there are two positions left after the i-th position,
// calculate the maximum stones Player2 can collect from those
// two positions
else if (i + 2 == n) {
c = arr[i] + arr[i + 1] - dp[i + 3];
}
// If there are more than two positions left after the i-th position,
// calculate the maximum stones Player2 can collect from the next three positions
if (b == -1)
b = arr[i] + arr[i + 1] - dp[i + 2];
if (c == -1)
c = arr[i] + arr[i + 1] + arr[i + 2] - dp[i + 3];
// Calculate the maximum stones that Player1 can collect starting from the i-th position
dp[i] = Math.Max(a, Math.Max(b, c));
}
// Determine the winner based on the maximum stones Player1
// can collect from the first position
if (dp[0] > 0)
return "Player1";
else if (dp[0] < 0)
return "Player2";
else
return "Tie";
}
public static void Main(string[] args) {
int[] arr = { 1, 2, 3, 7 };
int n = arr.Length;
Console.WriteLine(MaximumStones(arr, n));
}
}
// Function to find the winner of the game
function maximumStones(arr) {
let n = arr.length;
// Create an array dp of size N+3 and initialize it with 0
let dp = Array(n + 3).fill(0);
// Initialize the table for the last 3 stones
dp[n] = dp[n + 1] = dp[n + 2] = 0;
// Calculate the table for the remaining stones in reverse order
for (let i = n - 1; i >= 0; i--) {
// Pick one stone
let a = arr[i] - dp[i + 1];
let b = c = -1;
if (i + 1 === n) {
b = arr[i] - dp[i + 2];
c = arr[i] - dp[i + 3];
} else if (i + 2 === n) {
c = arr[i] + arr[i + 1] - dp[i + 3];
}
if (b === -1) {
b = arr[i] + arr[i + 1] - dp[i + 2];
}
if (c === -1) {
c = arr[i] + arr[i + 1] + arr[i + 2] - dp[i + 3];
}
dp[i] = Math.max(a, Math.max(b, c));
}
// Player 1 wins
if (dp[0] > 0) {
return "Player1";
}
// Player 2 wins
else if (dp[0] < 0) {
return "Player2";
}
// Tie
else {
return "Tie";
}
}
// Given Input
let arr = [1, 2, 3, 7];
// Function Call
console.log(maximumStones(arr));
// This code is contributed by srinivasteja
Time Complexity: O(N)
Auxiliary Space: O(N)
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