# Sort a 2D vector diagonally

Given a 2D vector of NxM integers. The task is to sort the elements of the vectors diagonally from top-left to bottom-right in decreasing order.

Examples:

Input: arr[][] = { { 10, 2, 3 }, { 4, 5, 6 }, {7, 8, 9 } }
Output:
10 2 0
4 9 0
0 0 5

Input: arr[][] = { { 10, 2, 43 }, { 40, 5, 16 }, { 71, 8, 29 }, {1, 100, 5} }
Output:
29 2 0
8 10 0
71 5 5
0 0 0

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:
Observations:

The above images shows the difference between the column index and row index at each cell. The cells having same difference from top-left to bottom-down cell forms a diagonal.
Below are the steps to sort diagonal in decreasing order:

1. Store the diagonal element with positive difference in one Array of Vectors(say Pos[]) such that elements at the cell having difference(say a) is stored at index a of Pos[] array.
2. Store the diagonal element with negative difference in another Array of Vectors(say Neg[]) such that elements at the cell having difference(say -b) is stored at index abs(-b) = b of Neg[] array.
3. Sort both the Array of Vectors increasing order.
4. Traverse the given 2D vector and updated the value at current cell with value stored in Pos[] and Neg[] array.
• If difference between column and row index(say d) is positive, then updated the value from Pos[d] array and remove the last elment as:
```d = i - j
arr[i][j] = Pos[d][Pos.size()-1]
Pos[d].pop_back()
```
• If difference between column and row index(say d) is negative, then updated the value from Neg[d] array and remove the last element as:
```d = j - i
arr[i][j] = Neg[d][Neg.size()-1]
Neg[d].pop_back()
```

Below is the implementation of the above approach:

 `// C++ program to sort the 2D vector ` `// diagonally in decreasing order ` `#include "bits/stdc++.h" ` `using` `namespace` `std; ` ` `  `// Function that sort the elements ` `// of 2D vector ` `void` `diagonalSort(vector >& mat) ` `{ ` ` `  `    ``// Calculate the rows and column ` `    ``int` `row = mat.size(); ` `    ``int` `col = mat.size(); ` ` `  `    ``// Array of vectors to store the ` `    ``// diagonal elements ` `    ``vector<``int``> Neg[row]; ` `    ``vector<``int``> Pos[col]; ` ` `  `    ``// Traverse the 2D vector and put ` `    ``// element in Array of vectors at ` `    ``// index difference between indexes ` `    ``for` `(``int` `i = 0; i < row; i++) { ` `        ``for` `(``int` `j = 0; j < col; j++) { ` ` `  `            ``// If diff is negative, then ` `            ``// push element to Neg[] ` `            ``if` `(j < i) { ` `                ``Neg[i - j].push_back(mat[i][j]); ` `            ``} ` ` `  `            ``// If diff is positive, then ` `            ``// push element to Pos[] ` `            ``else` `if` `(j > i) { ` `                ``Pos[j - i].push_back(mat[i][j]); ` `            ``} ` ` `  `            ``// If diff is 0, then push ` `            ``// element to Pos ` `            ``else` `{ ` `                ``Pos.push_back(mat[i][j]); ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Sort the Array of vectors ` `    ``for` `(``int` `i = 0; i < row; i++) { ` `        ``sort(Neg[i].begin(), Neg[i].end()); ` `    ``} ` `    ``for` `(``int` `i = 0; i < col; i++) { ` `        ``sort(Pos[i].begin(), Pos[i].end()); ` `    ``} ` ` `  `    ``// Update the value to arr[][] ` `    ``// from the sorted Array of vectors ` `    ``for` `(``int` `i = 0; i < row; i++) { ` `        ``for` `(``int` `j = 0; j < col; j++) { ` ` `  `            ``// If diff is positive ` `            ``if` `(j < i) { ` `                ``int` `d = i - j; ` `                ``int` `l = Neg[d].size() - 1; ` `                ``mat[i][j] = Neg[d][l - 1]; ` `                ``Neg[d].pop_back(); ` `            ``} ` ` `  `            ``// If diff is negative ` `            ``else` `if` `(j > i) { ` `                ``int` `d = j - i; ` `                ``int` `l = Pos[d].size() - 1; ` `                ``mat[i][j] = Pos[d][l - 1]; ` `                ``Pos[d].pop_back(); ` `            ``} ` ` `  `            ``// If diff is 0 ` `            ``else` `{ ` `                ``int` `l = Pos.size(); ` `                ``mat[i][j] = Pos[l - 1]; ` `                ``Pos.pop_back(); ` `            ``} ` `        ``} ` `    ``} ` `} ` ` `  `// Function to print element ` `void` `printElement(vector >& arr) ` `{ ` ` `  `    ``// Traverse the 2D vector ` `    ``for` `(``int` `i = 0; i < arr.size(); i++) { ` `        ``for` `(``int` `j = 0; j < arr.size(); j++) { ` `            ``cout << arr[i][j] << ``' '``; ` `        ``} ` `        ``cout << endl; ` `    ``} ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``vector > arr = { { 10, 2, 3 }, ` `                                 ``{ 4, 5, 6 }, ` `                                 ``{ 7, 8, 9 } }; ` `    ``diagonalSort(arr); ` ` `  `    ``// Function call to print elements ` `    ``printElement(arr); ` `} `

Output:
```10 2 0
4 9 0
0 0 5
```

Time Complexity: O(N2)

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