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 6 3
8 9 2
7 4 5
Input: arr[][] = { { 10, 2, 43 }, { 40, 5, 16 }, { 71, 8, 29 }, {1, 100, 5} }
Output:
29 16 43
40 10 2
100 8 5
1 71 5
Approach:
Observations:
The above images show the difference between the column index and row index at each cell. The cells having the same difference from top-left to bottom-down cell forms a diagonal.
Below are the steps to sort diagonal in decreasing order:
- Store the diagonal element with a positive difference in one Array of Vectors(say Pos[]) such that elements at the cell having difference(say a) is stored at index an of Pos[] array.
- Store the diagonal element with the 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.
- Sort both the Array of Vectors increasing order.
- Traverse the given 2D vector and updated the value at the current cell with the value stored in Pos[] and Neg[] array.
- If the difference between column and row index(say d) is positive, then updated the value from Pos[d] array and remove the last element as:
- If the difference between column and row index(say d) is positive, then updated the value from Pos[d] array and remove the last element as:
d = i - j arr[i][j] = Pos[d][Pos.size()-1] Pos[d].pop_back()
- If the 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:
CPP
// 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<vector< int > >& mat) { // Calculate the rows and column int row = mat.size(); int col = mat[0].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[0] else { Pos[0].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(); 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(); mat[i][j] = Pos[d][l - 1]; Pos[d].pop_back(); } // If diff is 0 else { int l = Pos[0].size(); mat[i][j] = Pos[0][l - 1]; Pos[0].pop_back(); } } } } // Function to print element void printElement(vector<vector< int > >& arr) { // Traverse the 2D vector for ( int i = 0; i < arr.size(); i++) { for ( int j = 0; j < arr[0].size(); j++) { cout << arr[i][j] << ' ' ; } cout << endl; } } // Driver Code int main() { vector<vector< int > > arr = { { 10, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } }; diagonalSort(arr); // Function call to print elements printElement(arr); } |
Java
// Java program to sort the 2D matrix // diagonally in decreasing order import java.io.*; import java.util.*; class GFG { public static void diagonalSort(ArrayList<ArrayList<Integer> > mat) { // Calculate the rows and column int row = mat.size(); int col = mat.get( 0 ).size(); // Arraylist of Arraylist to store the // diagonal elements ArrayList<ArrayList<Integer> > Neg = new ArrayList<ArrayList<Integer> >(); ArrayList<ArrayList<Integer> > Pos = new ArrayList<ArrayList<Integer> >(); int i, j; for (i = 0 ; i < row; i++) { ArrayList<Integer> temp = new ArrayList<Integer>(); Neg.add(temp); } for (j = 0 ; j < col; j++) { ArrayList<Integer> temp = new ArrayList<Integer>(); Pos.add(temp); } // Traverse the 2D matrix and put // element in Arraylist of Arraylist at // index difference between indexes for (i = 0 ; i < row; i++) { for (j = 0 ; j < col; j++) { // If diff is negative, then // push element to Neg[] if (j < i) { Neg.get(i - j).add(mat.get(i).get(j)); } // If diff is positive, then // push element to Pos[] else if (i < j) { Pos.get(j - i).add(mat.get(i).get(j)); } // If diff is 0, then push // element to Pos[0] else { Pos.get( 0 ).add(mat.get(i).get(j)); } } } // Sort the Array of vectors for (i = 0 ; i < row; i++) { Collections.sort(Neg.get(i)); ; } for (i = 0 ; i < col; i++) { Collections.sort(Pos.get(i)); ; } // Update the value to mat // from the sorted Arraylist of Arraylist for (i = 0 ; i < row; i++) { for (j = 0 ; j < col; j++) { // If diff is positive if (j < i) { int d = i - j; int l = Neg.get(d).size(); mat.get(i).set(j, Neg.get(d).get(l - 1 )); Neg.get(d).remove(l - 1 ); } // If diff is negative else if (i < j) { int d = j - i; int l = Pos.get(d).size(); mat.get(i).set(j, Pos.get(d).get(l - 1 )); Pos.get(d).remove(l - 1 ); } // If diff is 0 else { int l = Pos.get( 0 ).size(); mat.get(i).set(j, Pos.get( 0 ).get(l - 1 )); Pos.get( 0 ).remove(l - 1 ); } } } // Print diagonally sorted matrix for (i = 0 ; i < row; i++) { for (j = 0 ; j < col; j++) { System.out.print(mat.get(i).get(j) + " " ); } System.out.println(); } } // Driver Code public static void main(String[] args) { ArrayList<ArrayList<Integer> > arr = new ArrayList<ArrayList<Integer> >(); ArrayList<Integer> row1 = new ArrayList<Integer>(); row1.add( 10 ); row1.add( 2 ); row1.add( 3 ); arr.add(row1); ArrayList<Integer> row2 = new ArrayList<Integer>(); row2.add( 4 ); row2.add( 5 ); row2.add( 6 ); arr.add(row2); ArrayList<Integer> row3 = new ArrayList<Integer>(); row3.add( 7 ); row3.add( 8 ); row3.add( 9 ); arr.add(row3); diagonalSort(arr); } } // This code is contributed by Snigdha Patil |
Python3
# Python program for the above approach from collections import defaultdict def diagonalSort(matrix, n, m): # make a dict of list, where we # will store the diagonal elements to = defaultdict( list ) # store the diagonal elements with # respect to their row-col value # remember every row-col value for # each diagonal will be different for row in range (n): for col in range (m): to[row - col].append(matrix[row][col]) # sort the elements of each # diagonal as required for i in to: to[i].sort(reverse = True ) # store the new diagonal elements to # their respective position in the matrix for row in range (n): for col in range (m): matrix[row][col] = to[row - col].pop( 0 ) return matrix # Driver Code if __name__ = = "__main__" : matrix = [[ 10 , 2 , 3 ], [ 4 , 5 , 6 ], [ 7 , 8 , 9 ]] n = len (matrix) m = len (matrix[ 0 ]) matrix = diagonalSort(matrix, n, m) for row in range (n): for col in range (m): print (matrix[row][col], end = ' ' ) print () # This code is contributed by ajaymakvana. |
Javascript
// Javascript Program for the above approach function diagonalSort(matrix, n, m) { // create an object to store diagonal elements let to = {}; // store the diagonal elements with respect to their row-col value for (let row = 0; row < n; row++) { for (let col = 0; col < m; col++) { if (to[row - col] === undefined) { to[row - col] = []; } to[row - col].push(matrix[row][col]); } } // sort the elements of each diagonal as required for (let i in to) { to[i].sort((a, b) => b - a); } // store the new diagonal elements to their respective position in the matrix for (let row = 0; row < n; row++) { for (let col = 0; col < m; col++) { matrix[row][col] = to[row - col].shift(); } } return matrix; } // Driver Code let matrix = [ [10, 2, 3], [4, 5, 6], [7, 8, 9] ]; let n = matrix.length; let m = matrix[0].length; matrix = diagonalSort(matrix, n, m); for (let row = 0; row < n; row++) { let arr = []; for (let col = 0; col < m; col++) { arr.push(matrix[row][col]); } console.log(arr.join( " " )); } // This code is contributed princekumaras |
C#
using System; using System.Collections.Generic; using System.Linq; class GFG { // Driver's code static void Main( string [] args) { List<List< int > > arr = new List<List< int > >() { new List< int >{ 10, 2, 3 }, new List< int >{ 4, 5, 6 }, new List< int > { 7, 8, 9 } }; DiagonalSort(arr); // Function call to print elements PrintElement(arr); } // Function that sort the elements // of 2D vector static void DiagonalSort(List<List< int > > mat) { // Calculate the rows and column int row = mat.Count; int col = mat[0].Count; // Array of lists to store the diagonal elements List< int >[] Neg = new List< int >[ row ]; List< int >[] Pos = new List< int >[ col ]; for ( int i = 0; i < row; i++) { Neg[i] = new List< int >(); } for ( int i = 0; i < col; i++) { Pos[i] = new List< int >(); } // 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].Add(mat[i][j]); } // If diff is positive, then // push element to Pos[] else if (j > i) { Pos[j - i].Add(mat[i][j]); } // If diff is 0, then push // element to Pos[0] else { Pos[0].Add(mat[i][j]); } } } // Sort the Array of vectors for ( int i = 0; i < row; i++) { Neg[i].Sort(); } for ( int i = 0; i < col; i++) { Pos[i].Sort(); } // 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].Count; mat[i][j] = Neg[d][l - 1]; Neg[d].RemoveAt(l - 1); } // If diff is negative else if (j > i) { int d = j - i; int l = Pos[d].Count; mat[i][j] = Pos[d][l - 1]; Pos[d].RemoveAt(l - 1); } // If diff is 0 else { int l = Pos[0].Count; mat[i][j] = Pos[0][l - 1]; Pos[0].RemoveAt(l - 1); } } } } // Function to print element static void PrintElement(List<List< int > > arr) { // Traverse the 2D vector for ( int i = 0; i < arr.Count; i++) { for ( int j = 0; j < arr[i].Count; j++) { Console.Write(arr[i][j] + " " ); } Console.WriteLine(); } } } |
10 6 3 8 9 2 7 4 5
Time Complexity: O(N*M*log(min(N,M)))
Space Complexity: O(N*M)
Method 2 :
here in this method, we will do space optimization in the above method. here we traverse matrix diagonal and store their values in the extra 1D array so for every diagonal we will need to store the maximum min(n,m) element in our 1D array so this is space optimization in the above solution
C++
// C++ program to sort the 2D vector // diagonally in decreasing order #include <bits/stdc++.h> using namespace std; // Function that sort 2D matrix Diagonally In Descending // order void diagonalSort(vector<vector< int > >& mat) { // Calculate the rows and column int n = mat.size(); int m = mat[0].size(); // 1D array for extra space vector< int > v; // start traversing from first row to nth row // where first row to nth row is first member of // diagonal for ( int row = 0; row < n; row++) { // take all diagonal element where first element is // mat[row][0] means left column of matrix for ( int j = 0, i = row; i < n && j < m; i++, j++) { v.push_back(mat[i][j]); } // sort element in reverse order because we need // decreasing order in diagonal sort(v.rbegin(), v.rend()); int t = 0; // putting this all values to matrix in descending // sorted order for ( int j = 0, i = row; i < n && j < m; i++, j++) { mat[i][j] = v[t++]; } v.clear(); } // start traversing from second column to mth column // where second column to mth column is first member of // diagonal note that here we can't start from first // column because it is already sorted by first row // processing for ( int col = 1; col < m; col++) { // take all diagonal element where first element is // mat[0][col] means first row of matrix for ( int j = col, i = 0; i < n && j < m; i++, j++) { v.push_back(mat[i][j]); } // sort element in reverse order because we need // decreasing order in diagonal sort(v.rbegin(), v.rend()); int t = 0; // putting this all values to matrix in descending // sorted order for ( int j = col, i = 0; i < n && j < m; i++, j++) { mat[i][j] = v[t++]; } v.clear(); } } // Function to print element void printElement(vector<vector< int > >& arr) { // Traverse the 2D vector for ( int i = 0; i < arr.size(); i++) { for ( int j = 0; j < arr[0].size(); j++) { cout << arr[i][j] << ' ' ; } cout << endl; } } // Driver Code int main() { vector<vector< int > > arr = { { 10, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } }; diagonalSort(arr); // Function call to print elements printElement(arr); } |
Java
// Java program to sort the 2D matrix // diagonally in decreasing order import java.io.*; import java.util.*; class GFG { // Function that sort 2D matrix Diagonally In Descending // order public static void diagonalSort(ArrayList<ArrayList<Integer> > mat) { // Calculate the rows and column int n = mat.size(); int m = mat.get( 0 ).size(); // 1D array for extra space ArrayList<Integer> v = new ArrayList<Integer>(); // start traversing from first row to nth row // where first row to nth row is first member of // diagonal for ( int row = 0 ; row < n; row++) { // take all diagonal element where first element // is mat[row][0] means left column of matrix for ( int j = 0 , i = row; j < m && i < n; i++, j++) { v.add(mat.get(i).get(j)); } // sort element in reverse order because we need // decreasing order in diagonal Collections.sort(v, Collections.reverseOrder()); int t = 0 ; for ( int j = 0 , i = row; j < m && i < n; i++, j++) { mat.get(i).set(j, v.get(t++)); } v.clear(); } // start traversing from second column to mth column // where second column to mth column is first member // of diagonal note that here we can't start from // first column because it is already sorted by // first row processing for ( int col = 1 ; col < m; col++) { // take all diagonal element where first element // is mat[0][col] means first row of matrix for ( int j = col, i = 0 ; i < n && j < m; i++, j++) { v.add(mat.get(i).get(j)); } // sort element in reverse order because we need // decreasing order in diagonal Collections.sort(v, Collections.reverseOrder()); int t = 0 ; // putting this all values to matrix in // descending sorted order for ( int j = col, i = 0 ; i < n && j < m; i++, j++) { mat.get(i).set(j, v.get(t++)); } v.clear(); } // Print diagonally sorted matrix for ( int i = 0 ; i < n; i++) { for ( int j = 0 ; j < m; j++) { System.out.print(mat.get(i).get(j) + " " ); } System.out.println(); } } // Driver Code public static void main(String[] args) { ArrayList<ArrayList<Integer> > arr = new ArrayList<ArrayList<Integer> >(); ArrayList<Integer> row1 = new ArrayList<Integer>(); row1.add( 10 ); row1.add( 2 ); row1.add( 3 ); arr.add(row1); ArrayList<Integer> row2 = new ArrayList<Integer>(); row2.add( 4 ); row2.add( 5 ); row2.add( 6 ); arr.add(row2); ArrayList<Integer> row3 = new ArrayList<Integer>(); row3.add( 7 ); row3.add( 8 ); row3.add( 9 ); arr.add(row3); diagonalSort(arr); } } // This code is contributed by Snigdha Patil |
Python3
# Python program to sort the 2D vector diagonally in decreasing order # Function that sort 2D matrix Diagonally In Descending order def DiagonalSort(mat): # Calculate the rows and column n = len (mat) m = len (mat[ 0 ]) # 1D array for extra space v = [] # start traversing from first row to nth row # where first row to nth row is first member of diagonal for row in range ( 0 , n): j = 0 i = row # take all diagonal element where first element is # mat[row][0] means left column of matrix while (i < n and j < m): v.append(mat[i][j]) i + = 1 j + = 1 # sort element in reverse order because we need # decreasing order in diagonal v.sort(reverse = True ) v[:: - 1 ] t = 0 j = 0 i = row # putting this all values to matrix in descending sorted order while (i < n and j < m): mat[i][j] = v[t] t + = 1 i + = 1 j + = 1 v = [] # start traversing from second column to mth column # where second column to mth column is first member of diagonal # note that here we can't start from first column # because it is already sorted by first row processing for col in range ( 0 , m): j = col i = 0 # take all diagonal element where first element is # mat[0][col] means first row of matrix while (i < n and j < m): v.append(mat[i][j]) i + = 1 j + = 1 # sort element in reverse order because we need # decreasing order in diagonal v.sort(reverse = True ) v[:: - 1 ] t = 0 j = col i = 0 # putting this all values to matrix in descending sorted order while (i < n and j < m): mat[i][j] = v[t] t + = 1 i + = 1 j + = 1 v = [] return mat # Function to print element def printElement(arr): n = len (arr) m = len (arr[ 0 ]) # Traverse the 2D array for i in range ( 0 , n): for j in range ( 0 , m): print (arr[i][j], end = " " ) print () # Driver Code arr = [[ 10 , 2 , 3 ], [ 4 , 5 , 6 ], [ 7 , 8 , 9 ]] DiagonalSort(arr) printElement(arr) |
Javascript
// Function that sorts 2D matrix diagonally in decreasing order function diagonalSort(mat) { // Calculate the rows and columns const n = mat.length; const m = mat[0].length; // 1D array for extra space let v = []; // start traversing from first row to nth row // where first row to nth row is first member of diagonal for (let row = 0; row < n; row++) { // take all diagonal element where first element is // mat[row][0] means left column of matrix for (let j = 0, i = row; i < n && j < m; i++, j++) { v.push(mat[i][j]); } // sort element in reverse order because we need // decreasing order in diagonal v.sort((a, b) => b - a); let t = 0; // putting this all values to matrix in descending sorted order for (let j = 0, i = row; i < n && j < m; i++, j++) { mat[i][j] = v[t++]; } v = []; } // start traversing from second column to mth column // where second column to mth column is first member of diagonal // note that here we can't start from first column // because it is already sorted by first row processing for (let col = 1; col < m; col++) { // take all diagonal element where first element is // mat[0][col] means first row of matrix for (let j = col, i = 0; i < n && j < m; i++, j++) { v.push(mat[i][j]); } // sort element in reverse order because we need // decreasing order in diagonal v.sort((a, b) => b - a); let t = 0; // putting this all values to matrix in descending sorted order for (let j = col, i = 0; i < n && j < m; i++, j++) { mat[i][j] = v[t++]; } v = []; } } // Function to print elements function printElement(arr) { // Traverse the 2D array for (let i = 0; i < arr.length; i++) { let rowStr = ' '; for (let j = 0; j < arr[0].length; j++) { rowStr += arr[i][j] + ' '; } console.log(rowStr); } } // Driver Code const arr = [[10, 2, 3], [4, 5, 6], [7, 8, 9]]; diagonalSort(arr); // Function call to print elements printElement(arr); // This code has been contributed by Prince kumar |
C#
// C# program to sort the 2D vector diagonally in decreasing // order // Function that sort 2D matrix Diagonally In Descending // order using System; class Program { static void DiagonalSort( int [, ] mat) { // Calculate the rows and column int n = mat.GetLength(0); int m = mat.GetLength(1); // 1D array for extra space int [] v = new int [(Math.Max(n, m))]; // start traversing from first row to nth row // where first row to nth row is first member of // diagonal for ( int row = 0; row < n; row++) { int j = 0; int i = row; // take all diagonal element where first element // is mat[row][0] means left column of matrix while (i < n && j < m) { v[j] = mat[i, j]; i++; j++; } // sort element in reverse order because we need // decreasing order in diagonal Array.Sort(v, 0, j); Array.Reverse(v, 0, j); int t = 0; j = 0; i = row; // putting this all values to matrix in // descending sorted order while (i < n && j < m) { mat[i, j] = v[t]; t++; i++; j++; } } // start traversing from second column to mth column // where second column to mth column is first member // of diagonal note that here we can't start from // first column because it is already sorted by // first row processing for ( int col = 0; col < m; col++) { int j = col; int i = 0; // take all diagonal element where first element // is mat[0][col] means first row of matrix while (i < n && j < m) { v[i] = mat[i, j]; i++; j++; } // sort element in reverse order because we need // decreasing order in diagonal Array.Sort(v, 0, i); Array.Reverse(v, 0, i); int t = 0; j = col; i = 0; // putting this all values to matrix in // descending sorted order while (i < n && j < m) { mat[i, j] = v[t]; t++; i++; j++; } } } // Function to print element static void PrintElement( int [, ] arr) { int n = arr.GetLength(0); int m = arr.GetLength(1); // Traverse the 2D array for ( int i = 0; i < n; i++) { for ( int j = 0; j < m; j++) { Console.Write(arr[i, j] + " " ); } Console.WriteLine(); } } // Driver Code static void Main( string [] args) { int [, ] arr = { { 10, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } }; DiagonalSort(arr); PrintElement(arr); } } // This code is contributed by shiv1o43g |
10 6 3 8 9 2 7 4 5
Time Complexity: O(N*M*log(min(N,M)))
Auxiliary Space: O(min(N,M))
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