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# Sum of matrix element where each elements is integer division of row and column

• Difficulty Level : Medium
• Last Updated : 19 Aug, 2022

Consider a N X N matrix where each element is divided by a column number (integer division), i.e. mat[i][j] = floor((i+1)/(j+1)) where 0 <= i < n and 0 <= j < n. The task is to find the sum of all matrix elements.

Examples :

```Input  : N = 2
Output : 4
2 X 2 matrix with given constraint:
1 0
2 1
Sum of matrix element: 4

Input  : N = 3
Output : 9```

Method 1 (Brute Force): Run two loops, one for the row and another for the column, and find the integer part of (i / j) and add to the answer.
Below is the implementation of this approach:

## C++

 `// C++ program to find sum of matrix element` `// where each element is integer division of` `// row and column.` `#include` `using` `namespace` `std;`   `// Return sum of matrix element where each element` `// is division of its corresponding row and column.` `int` `findSum(``int` `n)` `{` `    ``int` `ans = 0;` `    ``for` `(``int` `i = 1; i <= n; i++)   ``// for rows` `        ``for` `(``int` `j = 1; j <= n; j++) ``// for columns` `            ``ans += (i/j);` `    ``return` `ans;` `}`   `// Driven Program` `int` `main()` `{` `    ``int` `N = 2;` `    ``cout << findSum(N) << endl;` `    ``return` `0;` `}`

## Java

 `// java program to find sum of matrix` `// element where each element is integer` `// division of row and column.`   `import` `java.io.*;`   `class` `GFG {` `    `  `    ``// Return sum of matrix element ` `    ``// where each element is division` `    ``// of its corresponding row and` `    ``// column.` `    ``static` `int` `findSum(``int` `n)` `    ``{` `        ``int` `ans = ``0``;` `        `  `        ``// for rows` `        ``for` `(``int` `i = ``1``; i <= n; i++) ` `        `  `            ``// for columns` `            ``for` `(``int` `j = ``1``; j <= n; j++) ` `                ``ans += (i/j);` `                `  `        ``return` `ans;` `    ``}` `    `  `    ``// Driven Program` `    ``public` `static` `void` `main (String[] args) ` `    ``{` `        ``int` `N = ``2``;` `        ``System.out.println( findSum(N));` `    ``}` `}`   `// This code is contributed by anuj_67.`

## Python3

 `# Python 3 program to find sum of ` `# matrix element where each element ` `# is integer division of row and column. `   `# Return sum of matrix element ` `# where each element is division ` `# of its corresponding row and column. ` `def` `findSum(N):` `    ``ans ``=` `0` `    ``for` `i ``in` `range``(``1``, N ``+` `1``):` `        ``for` `j ``in` `range``(``1``, N ``+` `1``):` `            ``ans ``+``=` `i ``/``/` `j` `    ``return` `ans`   `# Driver code` `N ``=` `2` `print``(findSum(N))`   `# This code is contributed ` `# by Shrikant13`

## C#

 `// C# program to find the sum of matrix` `// element where each element is an integer` `// division of row and column.` `using` `System;`   `class` `GFG {` `    `  `    ``// Return sum of matrix element ` `    ``// where each element is division` `    ``// of its corresponding row and` `    ``// column.` `    ``static` `int` `findSum(``int` `n)` `    ``{` `        ``int` `ans = 0;` `        `  `        ``// for rows` `        ``for` `(``int` `i = 1; i <= n; i++) ` `        `  `            ``// for columns` `            ``for` `(``int` `j = 1; j <= n; j++) ` `                ``ans += (i/j);` `                `  `        ``return` `ans;` `    ``}` `    `  `    ``// Driven Program` `    ``public` `static` `void` `Main () ` `    ``{` `        ``int` `N = 2;` `        ``Console.WriteLine( findSum(N));` `    ``}` `}`   `// This code is contributed by anuj_67.`

## PHP

 ``

## Javascript

 ``

Output

`4`

Time complexity: O(n2).
Auxiliary Space: O(1)

Method 2 (Efficient):

Let N = 9, the matrix will be ```Observe, for each jth column
mat[i][k] = 0, for 1 <= k < j, 1 <= i <= N
mat[i][k] = 1, for j <= k < 2*j, 1 <= i <= N
mat[i][k] = 2, for 2*j <= k < 3*j, 1 <= i <= N
and so on.
So, in each column i, there are i - 1 zero,
followed by i times 1, followed by i times 2, and so on.
We traverse matrix column by column and sum elements.```

Below is the implementation of this approach.

## C++

 `// C++ program to find sum of matrix element` `// where each element is integer division of` `// row and column.` `#include` `using` `namespace` `std;`   `// Return sum of matrix element where each` `// element is division of its corresponding` `// row and column.` `int` `findSum(``int` `n)` `{` `    ``int` `ans = 0, temp = 0, num;`   `    ``// For each column.` `    ``for` `(``int` `i = 1; i <= n && temp < n; i++)` `    ``{` `        ``// count the number of elements of` `        ``// each column. Initialize to i -1` `        ``// because number of zeroes are i - 1.` `        ``temp = i - 1;`   `        ``// For multiply` `        ``num = 1;`   `        ``while` `(temp < n)` `        ``{` `            ``if` `(temp + i <= n)` `                ``ans += (i * num);` `            ``else` `                ``ans += ((n - temp) * num);`   `            ``temp += i;` `            ``num ++;` `        ``}` `    ``}`   `    ``return` `ans;` `}`   `// Driven Program` `int` `main()` `{` `    ``int` `N = 2;` `    ``cout << findSum(N) << endl;` `    ``return` `0;` `}`

## Java

 `// java program to find sum of matrix element` `// where each element is integer division of` `// row and column.`   `import` `java.io.*;`   `class` `GFG {` `    `  `    ``// Return sum of matrix element where each` `    ``// element is division of its corresponding` `    ``// row and column.` `    ``static` `int` `findSum(``int` `n)` `    ``{` `        ``int` `ans = ``0``, temp = ``0``, num;` `    `  `        ``// For each column.` `        ``for` `(``int` `i = ``1``; i <= n && temp < n; i++)` `        ``{` `            `  `            ``// count the number of elements of` `            ``// each column. Initialize to i -1` `            ``// because number of zeroes are i - 1.` `            ``temp = i - ``1``;` `    `  `            ``// For multiply` `            ``num = ``1``;` `    `  `            ``while` `(temp < n)` `            ``{` `                ``if` `(temp + i <= n)` `                    ``ans += (i * num);` `                ``else` `                    ``ans += ((n - temp) * num);` `    `  `                ``temp += i;` `                ``num ++;` `            ``}` `        ``}` `    `  `        ``return` `ans;` `    ``}` `    `  `    ``// Driven Program` `    ``public` `static` `void` `main (String[] args) ` `    ``{` `        ``int` `N = ``2``;` `        ``System.out.println(findSum(N));` `    ``}` `}`   `// This code is contributed by anuj_67.`

## Python3

 `# Program to find sum of matrix element ` `# where each element is integer division  ` `# of row and column. `   `# Return sum of matrix element where each ` `# element is division of its corresponding ` `# row and column. ` `def` `findSum(n):` `    ``ans ``=` `0``; temp ``=` `0``;`   `    ``for` `i ``in` `range``(``1``, n ``+` `1``):`   `        ``# count the number of elements of ` `        ``# each column. Initialize to i -1 ` `        ``# because number of zeroes are i - 1. ` `        ``if` `temp < n:` `            ``temp ``=` `i ``-` `1`   `            ``# For multiply` `            ``num ``=` `1` `            ``while` `temp < n:` `                ``if` `temp ``+` `i <``=` `n:` `                    ``ans ``+``=` `i ``*` `num` `                ``else``:` `                    ``ans ``+``=` `(n ``-` `temp) ``*` `num` `                ``temp ``+``=` `i` `                ``num ``+``=` `1` `    ``return` `ans`   `# Driver Code` `N ``=` `2` `print``(findSum(N))`   `# This code is contributed by Shrikant13`

## C#

 `// C# program to find sum of matrix ` `// element where each element is ` `// integer division of row and column.` `using` `System;`   `class` `GFG ` `{` `    `  `    ``// Return sum of matrix element ` `    ``// where each element is division ` `    ``// of its corresponding row and column.` `    ``static` `int` `findSum(``int` `n)` `    ``{` `        ``int` `ans = 0, temp = 0, num;` `    `  `        ``// For each column.` `        ``for` `(``int` `i = 1; i <= n && temp < n; i++)` `        ``{` `            `  `            ``// count the number of elements ` `            ``// of each column. Initialize ` `            ``// to i -1 because number of ` `            ``// zeroes are i - 1.` `            ``temp = i - 1;` `    `  `            ``// For multiply` `            ``num = 1;` `    `  `            ``while` `(temp < n)` `            ``{` `                ``if` `(temp + i <= n)` `                    ``ans += (i * num);` `                ``else` `                    ``ans += ((n - temp) * num);` `    `  `                ``temp += i;` `                ``num ++;` `            ``}` `        ``}` `    `  `        ``return` `ans;` `    ``}` `    `  `    ``// Driver Code` `    ``public` `static` `void` `Main () ` `    ``{` `        ``int` `N = 2;` `        ``Console.WriteLine(findSum(N));` `    ``}` `}`   `// This code is contributed by anuj_67.`

## PHP

 ``

## Javascript

 ``

Output

`3`

Time complexity: O(n2)
Auxiliary Space: O(1), since no extra space has been taken.

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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