Levenshtein to get edit distance between two strings (ignores newlines, but counts spaces)

Levenshtein

- (float)compareString:(NSString *)originalString withString:(NSString *)comparisonString {
    // Normalize strings
    [originalString stringByTrimmingCharactersInSet:NSCharacterSet.newlineCharacterSet];
    [comparisonString stringByTrimmingCharactersInSet:NSCharacterSet.newlineCharacterSet];
    
    originalString = originalString.lowercaseString;
    comparisonString = comparisonString.lowercaseString;
    
    // Step 1
    NSInteger k, i, j, cost, * d, distance;
    
    NSInteger n = originalString.length;
    NSInteger m = comparisonString.length;
    
    if (n == 0) {
        //edit distance is the entire length of the new string
        return m;
    }
    
    if( n++ != 0 && m++ != 0 ) {
        
        d = malloc( sizeof(NSInteger) * m * n );
        
        // Step 2
        for( k = 0; k < n; k++)
            d[k] = k;
        
        for( k = 0; k < m; k++)
            d[ k * n ] = k;
        
        // Step 3 and 4
        for( i = 1; i < n; i++ )
            for( j = 1; j < m; j++ ) {
                
                // Step 5
                if( [originalString characterAtIndex: i-1] ==
                   [comparisonString characterAtIndex: j-1] )
                    cost = 0;
                else
                    cost = 1;
                
                // Step 6
                d[ j * n + i ] = [self smallestOf: d [ (j - 1) * n + i ] + 1
                                            andOf: d[ j * n + i - 1 ] +  1
                                            andOf: d[ (j - 1) * n + i - 1 ] + cost ];
                
                // This conditional adds Damerau transposition to Levenshtein distance
                if( i>1 && j>1 && [originalString characterAtIndex: i-1] ==
                   [comparisonString characterAtIndex: j-2] &&
                   [originalString characterAtIndex: i-2] ==
                   [comparisonString characterAtIndex: j-1] )
                {
                    d[ j * n + i] = MIN(d[ j * n + i ], d[ (j - 2) * n + i - 2 ] + cost);
                }
            }
        
        distance = d[ n * m - 1 ];
        
        free(d);
        
        return distance;
    }
    return 0.0;
}

// Return the minimum of a, b and c - used by compareString:withString:
- (NSInteger)smallestOf:(NSInteger)a andOf:(NSInteger)b andOf:(NSInteger)c {
    return MIN(a, MIN(b, c));
}