max shows of number in two dimension array [closed] - c++

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I have a two dimensional array and I know:
The number of lines and the length of each line
Each line contains just positive numbers
Each line is sorted
not use with auxiliary array
-not use with data structures
Required Output
I need to return the number which appears max times in the the whole array in an efficient way.
I already tried to pass all over the array but it's not efficient.
This is an example of the array.
{
{5, 7, 8},
{6, 6},
{null},
{5, 6, 8, 9}
}
The expected return value for this example is 6.
I would like to get the explanation or code in c++
Thanks

Answer 1

Since a C/C++ solution is required then a 2D Array can be used.
All the missing values can be represented by -1 (or any number which is not expected in valid numbers involved in search).
So a empty row can be represented by all -1. See the code below.
Since in C/C++, 2D array is continuously represented in memory. So we can convert 2D array to 1D array.
Now we can sort the array. After sorting, all the '-1' will be at the beginning which can be discarded.
From the leftover elements we can find the max frequency of an element.
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int compare (const void * a, const void * b)
{
return ( *(int*)a - *(int*)b );
}
int main()
{
int i, prev, max = -1, count = 0, maxvalue = -1;
int a[4][4] = {{5, 7, 8, -1}, {6, 6, -1, -1}, {-1, -1, -1, -1}, {5, 6, 8, 9}};
//int a[4][4] = {{-1, -1, -1, -1}, {-1, -1, -1, -1}, {-1, -1, -1, -1}, {-1, -1, -1, -1}};
int *b = (int*)a;
int total = sizeof(a) / sizeof(int);
qsort(b, total, sizeof(int), compare);
for(i = 0; i < total; ++i)
{
if(b[i] != -1)
{
break;
}
}
//printf("\n");
i = i + 1;
prev = -1;
count = 0;
if(i < total)
{
prev = b[i];
count = 1;
}
for(i = i + 1; i < total; ++i)
{
//printf("prev=%d, b[i]=%d, max=%d, count=%d\n", prev, b[i], max, count);
if(prev == b[i])
{
count++;;
}
else
{
if(max < count)
{
max = count;
maxvalue = prev;
}
prev = b[i];
count = 1;
}
}
if(max != -1)
{
printf("Max Occurence of %d = %d\n", maxvalue, max);
}
else
{
printf("All the rows are of zero length\n");
}
return 0;
}
//Output:
Max Occurence of 6 = 3

Answer 2

For counting the number of times an element occurs in the array, a similar question using recursion is shown here.
Since you mentioned efficiency, it would help to sort the array in increasing or decreasing order prior to counting the number of times an element is present in the array (if unsorted). Although for a small input size as shown in your example, it wouldn't make much of a difference.

Answer 3

You could use map to keep track of the number of times its been repeated and a current max.
map<int, int> temp;
int currentMax= -999999,maxCount=0;
for(i=0; i< numberOflines ;i++)
{
for(j=0;j< array[i].length;j++)
{
int newCount = ++temp[array[i][j]];
if (maxCount < newCount) {
maxCount = newCount;
currentMax = array[i][j];
}
}
}

Answer 4

First off your input is illegal:
{
{5, 7, 8},
{6, 6},
{null},
{5, 6, 8, 9}
}
null is not defined by C++, and even if it was defined as 0 it would have to be interpreted as int(0), not an empty sub-array as I think that you intended.
I'm guessing that the input that you intend to imply should look something like this:
const initializer_list<int> a[] = {{5, 7, 8},
{6, 6},
{},
{5, 6, 8, 9}};
You'll need to maintain a total for each number in any array. The best way to do that is to use map<int, int> totals which will only be constructed with the exact number of pairs as there are unique elements in a. The second element of each pair will be the count of that element seen thus far. You can populate it by doing:
for(const auto& i : a) for_each(cbegin(i), cend(i), [&](const auto& it){ totals[it]++;});
Once totals is populated you need only find its largest value:
cout << max_element(cbegin(totals), cend(totals), [](const auto& lhs, const auto& rhs){return lhs.second < rhs.second;})->first << endl;
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