fixed indentation and blank lines

Code/C++/kadane_algo.cpp

@@ -1,99 +1,90 @@
 /* 
-Kadane's Algorithm is used to solve the maximum subarray sum problem.
+	Kadane's Algorithm is used to solve the maximum subarray sum problem.
 
-The maximum subarray problem is the task of finding the largest 
-possible sum of a contiguous subarray.
-	
-We make three cases in this algorithms.
+	The maximum subarray problem is the task of finding the largest 
+	possible sum of a contiguous subarray.
+
+	We make three cases in this algorithms.
 
-1 - all the array elements are postive - then complete array 
-sum is the largest subarray sum.
+	1 - all the array elements are postive - then complete array 
+	sum is the largest subarray sum.
 
-2 - all the array elemets are negative - then smallest elements 
-is the largest subarray sum.
+	2 - all the array elemets are negative - then smallest elements 
+	is the largest subarray sum.
 
-3 - elements are mix of positive and negative numbers - then we keep adding 
-elements in the variable until the variable is positive. We keep track of the
-variable value after each addition, so that we know the maximum value it rose to,
-that maximum value is our answer. 
+	3 - elements are mix of positive and negative numbers - then we keep adding 
+	elements in the variable until the variable is positive. We keep track of the
+	variable value after each addition, so that we know the maximum value it rose to,
+	that maximum value is our answer. 
 */
 
 #include <bits/stdc++.h>
 #define ll long long
 using namespace std;
+
 main() 
 {
 	ios_base::sync_with_stdio(false);
-    cin.tie(NULL);
+    		cin.tie(NULL);
 
 	ll int n,maxs;
 
 	// no of elements in the array
 	cin>>n;
-
-
+
 	ll int arr[n],i,sum=0,pos=0,neg=0, maxi = INT_MIN;
 
 	for(i=0;i<n;i++)
 	{
-
-		// array input
+	    // array input
 	    cin>>arr[i];
 
-		// complete sum of the array 
-		sum += arr[i];
+	    // complete sum of the array 
+	    sum += arr[i];
 
-		// no of positive values
+	    // no of positive values
 	    if(arr[i] >=0 )
-	    pos++;
+	    	pos++;
 
-		// no of negative values  
+	    // no of negative values  
 	    else
-	    neg++;
-
+	    	neg++;		
 	}
 
 	// if all elements are positive   
-	if(pos == n)
-
-	printf("%lld\n",sum);
+	if(pos == n)	
+		printf("%lld\n",sum);
 
 	// if all elements are negative   
-	else if(neg == n)
-
+	else if(neg == n)	
 	{
-		// find the largest element
+	    // find the largest element
 	    for(i = 0; i<n; i++)
 	    maxi = max(maxi , arr[i]);
 
-		// print the largest elements   
+            // print the largest elements   
 	    printf("%lld\n",maxi);
-
 	}
+
 	else
 	{
 	    ll int left = 0;
 	    ll int maxs = 0;
 
 	    for(i=0; i<n; i++)
-	    {
-
+	    {	    	
 	    	// if variable becomes negative, reset left to 0
-	        if( left + arr[i] <=0 )
-
-	        {        
+	        if( left + arr[i] <=0 )				      
 	            left = 0;
-	        }
-
+
 	        else
-				// add if the variable will remain positive
+		// add if the variable will remain positive
 	        {
 	            left += arr[i];
 	            maxs = max(maxs,left);
-				// keep track of variable's value to know the maximum
-
-	        }
+		    // keep track of variable's value to know the maximum
 
+	        }	            
 	    }
 
 	    // print the answer