From f60402f101b6af77db05c22a4a2bb8106e7a50fe Mon Sep 17 00:00:00 2001
From: Yashgabani845 <gabaniyash845@gmail.com>
Date: Sun, 2 Jun 2024 17:50:35 +0530
Subject: [PATCH 1/2] kadane's algorithm solution added

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 create mode 100644 dsa-solutions/gfg-solutions/0003-kadane's-algorithm.md

diff --git a/dsa-solutions/gfg-solutions/0003-kadane's-algorithm.md b/dsa-solutions/gfg-solutions/0003-kadane's-algorithm.md
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+---
+id: kadane's-algorithm
+title: Kadane's Alogrithm (Geeks for Geeks)
+sidebar_label: 0003 - Kadane's Algorithm 
+tags:
+  - intermediate
+  - Array
+  - Dynamic Programming
+  - Data Structure
+  - Algorithms
+
+description: "This is a solution to the Kadane's Algorithm problem on Geeks for Geeks."
+---
+
+This tutorial contains a complete walk-through of the Kadane's Algorithm  problem from the Geeks for Geeks website. It features the implementation of the solution code in three programming languages: Python ,C++ ,java.
+
+## Problem Description
+
+Given an array Arr[] of N integers, find the contiguous sub-array (containing at least one number) which has the maximum sum and return its sum.
+
+## Examples
+
+**Example 1:**
+
+```
+Input:
+N = 5
+Arr[] = {1,2,3,-2,5}
+Output:
+9
+Explanation:
+Max subarray sum is 9
+of elements (1, 2, 3, -2, 5) which 
+is a contiguous subarray.
+```
+
+**Example 2:**
+
+```
+Input:
+N = 4
+Arr[] = {-1,-2,-3,-4}
+Output:
+-1
+Explanation:
+Max subarray sum is -1 
+of element (-1)
+```
+
+## Your Task
+
+You don't need to read input or print anything. Your task is to complete the function `maxSubarraySum()` which takes Arr[] and N as input parameters and returns the sum of subarray with maximum sum.
+
+Expected Time Complexity: $O(N)$  
+Expected Auxiliary Space: $O(1)$
+
+## Constraints
+
+1. $(1 \leq N \leq 10^6)$
+2. $(-10^7 \leq A[i] \leq 10^7)$
+
+## Solution Approach
+
+### Brute Force Approach
+
+#### Intuition:
+Try all possible subarrays by using nested loops and pick the subarray which has the maximum sum.
+
+#### Implementation:
+1. Keep an answer variable to store the maximum subarray sum.
+2. Run a loop(i).
+3. Run a nested loop from i till the end of the array.
+4. Generate all subarrays starting from the ith index and compare its sum with the answer and update the answer with the maximum one.
+5. Return the answer.
+
+#### Code (C++):
+
+```cpp
+class Solution{
+    public:
+    // arr: input array
+    // n: size of array
+    //Function to find the sum of contiguous subarray with maximum sum.
+    long long maxSubarraySum(int arr[], int n){
+        
+        // Your code here
+        long long ans=arr[0];
+        for(int i=0;i<n;i++) {
+            long long temp=0;
+            for(int j=i;j<n;j++) {
+                temp += arr[j];
+                ans = max(ans,temp);
+            }
+        }
+        return ans;
+    }
+};
+```
+
+#### Complexity:
+- Time Complexity: $(O(N^2))$, As we are running a nested loop over the length of the array.
+- Space Complexity: $(O(1))$, Here no extra space is used.
+
+### Kadane's Algorithm
+
+#### Intuition:
+Kadane’s algorithm works on the principle of continuously adding the array value to the current sum so far and updating the max sum every time. Whenever the current sum becomes negative, initialize it to zero and continue the process. By this approach, only one loop is required to compute the maximum sum contiguous subarray, reducing the time complexity to linear $(O(N))$.
+
+#### Implementation: 
+1. Initialize the variables `max_so_far = arr[0]` (stores the maximum sum of contiguous subarray found so far) and `max_ending_here = 0` (stores the maximum sum contiguous subarray ending at the current index).
+2. Run a for loop from 0 to N-1 and for each index i: 
+    - Add the arr[i] to `max_ending_here`.
+    - If `max_so_far` is less than `max_ending_here`, then update `max_so_far` to `max_ending_here`.
+    - If `max_ending_here` < 0, then update `max_ending_here = 0`.
+3. Return `max_so_far`.
+
+#### Example:
+Let's take the example: {-2, -3, 4, -1, -2, 1, 5, -3}
+- `max_so_far = arr[0] = -2`
+- `max_ending_here = 0`
+
+For i=0, a[0] = -2:
+- `max_ending_here = max_ending_here + (-2)`
+- Set `max_ending_here = 0` because `max_ending_here < 0`
+- Set `max_so_far = -2`
+
+For i=1, a[1] = -3:
+- `max_ending_here = max_ending_here + (-3)`
+- Since `max_ending_here = -3` and `max_so_far = -2`, `max_so_far` will remain -2
+- Set `max_ending_here = 0` because `max_ending_here < 0`
+
+For i=2, a[2] = 4:
+- `max_ending_here = max_ending_here + (4)`
+- `max_ending_here = 4`
+- `max_so_far` is updated to 4 because `max_ending_here` greater than `max_so_far` which was -2 till now
+
+For i=3, a[3] = -1:
+- `max_ending_here = max_ending_here + (-1)`
+- `max_ending_here = 3`
+
+For i=4, a[4] = -2:
+- `max_ending_here = max_ending_here + (-2)`
+- `max_ending_here = 1`
+
+For i=5, a[5] = 1:
+- `max_ending_here = max_ending_here + (1)`
+- `max_ending_here = 2`
+
+For i=6, a[6] = 5:
+- `max_ending_here = max_ending_here + (5)`
+- `max_ending_here = 7`
+- `max_so_far` is updated to 7 because `max_ending_here` is greater than `max_so_far`
+
+For i=7, a[7] = -3:
+- `max_ending_here = max_ending_here + (-3)`
+- `max_ending_here = 4`
+
+Hence, `max_so_far = 7`.
+
+#### Code (C++):
+
+```cpp
+class Solution{
+    public:
+    //Function to find the sum of contiguous subarray with maximum sum.
+    long long maxSubarraySum(int a[], int n){
+        
+        long long maxh = 0, maxf = a[0];
+        
+        //Iterating over the array.
+        for(int i=0;i<n;i++) 
+            {
+                //Updating max sum till current index.
+                maxh+=a[i];
+                //Storing max sum so far by choosing maximum between max 
+                //sum so far and max till current index.
+                if(maxf<maxh)
+                maxf=maxh; 
+                
+                //If max sum at current index is negative, we do not need to add
+                //it to result so we update it to zero.
+                if(maxh<0)
+                    maxh=0;
+    
+            }
+            //returning the result.
+            return maxf;
+        
+    }
+};
+```
+
+#### Code (Java):
+
+```java
+class Solution {
+    //Function to find the sum of contiguous subarray with maximum sum.
+    long maxSubarraySum(int arr[], int n){
+        
+        long maxh = 0, maxf = arr[0];
+        
+        //Iterating over the array.
+        for(int i = 0; i < n; i++){
+            
+            //Updating max sum till current index.
+            maxh += arr[i];
+            //Storing max sum so far by choosing maximum between max 
+            //sum so far and max sum till current index.
+            if(maxf < maxh)
+                maxf = maxh; 
+            
+            //If max sum till current index is negative, we do not need to add
+            //it to result so we update it to zero.
+            if(maxh < 0)
+                maxh = 0;
+        }
+        //returning the result.   
+        return maxf;
+    }
+}
+```
+
+#### Code (Python):
+
+```python
+class Solution:
+    #Function to find the sum of contiguous subarray with maximum sum.
+    def maxSubArraySum(self,a,size): 
+           
+        max_so_far = -9999999 - 1
+        max_ending_here = 0
+        
+        #Iterating over the array. 
+        for i in range(0, size): 
+            #Updating max sum till current index.
+            max_ending_here = max_ending_here + a[i]
+            
+            #Storing max sum so far by choosing maximum between max 
+            #sum so far and max sum till current index.
+            if (max_so_far < max_ending_here): 
+                max_so_far = max_ending_here 
+        
+            #If max sum till current index is negative, we do not need to add
+            #it to result so we update it to zero.
+            if max_ending_here < 0: 
+                max_ending_here = 0   
+        
+        #returning the result.
+        return max_so_far
+```
+
+#### Complexity:
+- Time Complexity: $(O(N))$, where O(N)$ is the size of the array as we are looping once over the whole array.
+- Space Complexity: $(O(1))$, Here we are not using extra space.
+
+## Conclusion
+
+The problem of finding the maximum sum of a contiguous subarray can be effectively solved using Kadane's Algorithm, which operates in linear time $(O(N))$. This algorithm is highly efficient and utilizes constant space, making it suitable for large input sizes as specified in the constraints. The brute force approach, while conceptually simpler, is computationally expensive with a time complexity of $(O(N^2))$ and is not suitable for larger arrays.
+
+By iterating through the array and maintaining the current
+
+ subarray sum and the maximum subarray sum found so far, Kadane's Algorithm offers an optimal solution. This approach ensures that even in the worst case, where all elements are negative, the algorithm correctly identifies the largest element, which is the maximum sum subarray in such scenarios.
+
+The provided implementations in C++, Java, and Python demonstrate how Kadane's Algorithm can be applied across different programming languages, ensuring versatility and ease of integration into various codebases.
+
+## References
+
+- **GeeksforGeeks Problem:** [Geeks for Geeks Problem](https://www.geeksforgeeks.org/problems/kadanes-algorithm-1587115620/1)
+- **Solution Link:** [Kadane's Algorithm on Geeks for Geeks](https://www.geeksforgeeks.org/problems/kadanes-algorithm-1587115620/1)
+- **Authors GeeksforGeeks Profile:** [Yash](https://www.geeksforgeeks.org/user/gabaniyxn26/)
\ No newline at end of file

From d0569fd5b2f989195d295a7f8f28c4289c9d95da Mon Sep 17 00:00:00 2001
From: Yashgabani845 <gabaniyash845@gmail.com>
Date: Sun, 2 Jun 2024 17:52:50 +0530
Subject: [PATCH 2/2] complexity style added

---
 dsa-solutions/gfg-solutions/0003-kadane's-algorithm.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/dsa-solutions/gfg-solutions/0003-kadane's-algorithm.md b/dsa-solutions/gfg-solutions/0003-kadane's-algorithm.md
index c27a3ed9f..12ee2ecc8 100644
--- a/dsa-solutions/gfg-solutions/0003-kadane's-algorithm.md
+++ b/dsa-solutions/gfg-solutions/0003-kadane's-algorithm.md
@@ -250,7 +250,7 @@ class Solution:
 ```
 
 #### Complexity:
-- Time Complexity: $(O(N))$, where O(N)$ is the size of the array as we are looping once over the whole array.
+- Time Complexity: $(O(N))$, where $O(N)$ is the size of the array as we are looping once over the whole array.
 - Space Complexity: $(O(1))$, Here we are not using extra space.
 
 ## Conclusion