Just finish a fun question from Hackerrank: Sherlock and Array
Question description:

Watson gives Sherlock an array AA of length NN. Then he asks him to determine if there exists an element in the array such that the sum of the elements on its left is equal to the sum of the elements on its right. If there are no elements to the left/right, then the sum is considered to be zero.
Formally, find an ii, such that, AA1+A+A2…A…Ai-1 =A=Ai+1+A+Ai+2…A…AN.

My approach is to go through array and on each iteration adjust sum on the left and sum on the right by current integer from array. Complexity of this approach is O(n). Code below:

#import <Foundation/Foundation.h>

int main (int argc, const char * argv[]) {

    NSAutoreleasePool *pool = [[NSAutoreleasePool alloc] init];

    int numberOfTestCases;
    scanf("%d", &numberOfTestCases);
   
    for (int i=0; i<numberOfTestCases; i++) {
         NSMutableArray *inputIntegers = [NSMutableArray new];
        BOOL sumExists = NO; 
        int numberOfElements;
        int sumLeft = 0;
        int sumRight = 0;
        scanf("%d", &numberOfElements);
        for (int j=0; j<numberOfElements; j++) {
            int inputElement;
            scanf("%d", &inputElement);
            [inputIntegers addObject:[NSNumber numberWithInt:inputElement]];
            sumRight+=[inputIntegers[j] integerValue];
        }
       
        for (int j=0; j<numberOfElements; j++) {
            sumRight -= [inputIntegers[j] integerValue];
            if (sumLeft == sumRight){
                sumExists = YES;
                break;
            } else {
                sumLeft += [inputIntegers[j] integerValue];
            }
        }
        printf("%s\n", sumExists ? "YES" : "NO");
    }

);

    [pool drain];
    return 0;
}

This question was previously asked on the Facebook interview.
Given input as a binary tree print to console each level followed by new line character.
Input:
Output:
“A\n
BC\n
DEF”

Solution: Do a breadth first search algorithm on a graph which will go over an entire tree and will build a resulted string.
First of all let’s create a custom class Node:

@interface Node : NSObject

@property NSObject *left;
@property NSObject *right;
@property NSString *value;

@end

Second let’s extend NSMutableArray with our own queue methods:

@interface  NSMutableArray (QueueAdditions)
- (id) dequeue;
- (void) enqueue: (id)obj;
@end

@implementation  NSMutableArray (QueueAdditions)

- (id) dequeue {
    id headObject = [self objectAtIndex:0];
    if (headObject != nil) {
        [self removeObjectAtIndex:0];
    }
    return headObject;
}

- (void) enqueue:(id)obj{
    [self addObject:obj];
}
@end

Add the function that will traverse over the graph and will build a resulted string:

+(NSString*) printTree:(Node*)node{
    
    if (node == nil){
        return @"";
    }
    
    NSMutableString *result = [[NSMutableString alloc] init];
    NSMutableArray *queue = [[NSMutableArray alloc] init];
    [queue enqueue:node];
    [queue enqueue:[NSNull null]];
    while (true){
        Node *curObject = [queue dequeue];
        if ([curObject isEqual:[NSNull null]]){
            [result appendString:@"\n"];
            
            if ([queue count] == 0){
                break;
            }
            [queue addObject:[NSNull null]];
            continue;
        }
        [result appendString:curObject.value];
        
        if (curObject.left){
            [queue enqueue:curObject.left];
        }
        if (curObject.right){
            [queue enqueue:curObject.right];
        }
    }
    
    return result;
}

code at gitlab: https://github.com/svetdev/print-tree-by-level