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Zoekelement in een gesorteerde matrix

Gegeven een gesorteerde matrix samen met[][] met een grootte n × m en een geheel getal X bepaal of x aanwezig is in de matrix.
De matrix wordt op de volgende manier gesorteerd:

  • Elke rij wordt in oplopende volgorde gesorteerd.
  • Het eerste element van elke rij is groter dan of gelijk aan het laatste element van de vorige rij
    (dat wil zeggen mat[i][0] ≥ mat[i−1][m−1] voor alle 1 ≤ i< n).

Voorbeelden:

Invoer: x = 14 mat[][] = [[ 1 5 9]
[14 20 21]
[30 34 43]]
Uitgang: WAAR
Uitleg: De waarde14is aanwezig in de tweede rij, eerste kolom van de matrix.



Invoer: x = 42 mat[][] = [[ 1 5 9 11]
[14 20 21 26]
[30 34 43 50]]
Uitgang: vals
Uitleg: De waarde42komt niet voor in de matrix.

Inhoudsopgave

[Naïeve benadering] Vergelijking met alle elementen – O(n × m) Tijd en O(1) Ruimte

Het idee is om de hele matrixmat[][] te doorlopen en elk element met x te vergelijken. Als een element overeenkomt met de x, retourneren we true. Anders zullen we aan het einde van de doortocht vals terugkeren.

C++ 
#include    #include  using namespace std; bool searchMatrix(vector<vector<int>>& mat int x) {  int n = mat.size();  int m = mat[0].size();  // traverse every element in the matrix  for (int i = 0; i < n; i++) {  for (int j = 0; j < m; j++) {  if (mat[i][j] == x)  return true;  }  }  return false; } int main() {  vector<vector<int>> mat = {  {1 5 9}  {14 20 21}  {30 34 43}  };  int x = 14;  cout << (searchMatrix(mat x) ? 'true' : 'false') << endl; }
Java 
class GfG {  public static boolean searchMatrix(int[][] mat int x) {  int n = mat.length;  int m = mat[0].length;  // traverse every element in the matrix  for (int i = 0; i < n; i++) {  for (int j = 0; j < m; j++) {  if (mat[i][j] == x)  return true;  }  }  return false;  }  public static void main(String[] args) {  int[][] mat = {  {1 5 9}  {14 20 21}  {30 34 43}  };  int x = 14;  System.out.println(searchMatrix(mat x) ? 'true' : 'false');  } }
Python 
def searchMatrix(mat x): n = len(mat) m = len(mat[0]) # traverse every element in the matrix for i in range(n): for j in range(m): if mat[i][j] == x: return True return False if __name__ == '__main__': mat = [ [1 5 9] [14 20 21] [30 34 43] ] x = 14 print('true' if searchMatrix(mat x) else 'false')
C# 
using System; class GfG {  public static bool searchMatrix(int[][] mat int x) {  int n = mat.Length;  int m = mat[0].Length;  // traverse every element in the matrix  for (int i = 0; i < n; i++) {  for (int j = 0; j < m; j++) {  if (mat[i][j] == x)  return true;  }  }  return false;  }  public static void Main(string[] args) {  int[][] mat = new int[][] {  new int[] {1 5 9}  new int[] {14 20 21}  new int[] {30 34 43}  };  int x = 14;  Console.WriteLine(searchMatrix(mat x) ? 'true' : 'false');  } }
JavaScript 
function searchMatrix(mat x) {  let n = mat.length;  let m = mat[0].length;  // traverse every element in the matrix  for (let i = 0; i < n; i++) {  for (let j = 0; j < m; j++) {  if (mat[i][j] === x)  return true;  }  }  return false; } // Driver Code let mat = [  [1 5 9]  [14 20 21]  [30 34 43] ]; let x = 14; console.log(searchMatrix(mat x) ? 'true' : 'false');

Uitvoer 
true

[Betere aanpak] Tweemaal binair zoeken gebruiken - O(log n + log m) Tijd en O(1) Ruimte

Eerst lokaliseren we de rij waar doel x zich zou kunnen bevinden door binair zoeken te gebruiken en vervolgens passen we binair zoeken opnieuw toe binnen die rij. Om de juiste rij te vinden, voeren we een binaire zoekopdracht uit op de eerste elementen van de middelste rij.

Stapsgewijze implementaties:

=> Begin met laag = 0 en hoog = n - 1.
=> Als x kleiner is dan het eerste element van de middelste rij (a[mid][0]), dan zal x kleiner zijn dan alle elementen in rijen >= mid, dus update hoog = mid - 1.
=> Als x groter is dan het eerste element van de middelste rij (a[mid][0]), dan zal x groter zijn dan alle elementen in rijen< mid so store the current mid row and update low = mid + 1.

Zodra we de juiste rij hebben gevonden, kunnen we binnen die rij binair zoeken om naar het doelelement x te zoeken.

C++ 
#include    #include  using namespace std; // function to binary search for x in arr[] bool search(vector<int> &arr int x) {  int lo = 0 hi = arr.size() - 1;  while (lo <= hi) {  int mid = (lo + hi) / 2;  if (x == arr[mid])  return true;  if (x < arr[mid])  hi = mid - 1;  else  lo = mid + 1;  }  return false; } // function to search element x in fully  // sorted matrix bool searchMatrix(vector<vector<int>> &mat int x) {  int n = mat.size() m = mat[0].size();  int lo = 0 hi = n - 1;  int row = -1;  while (lo <= hi) {  int mid = (lo + hi) / 2;  // if the first element of mid row is equal to x  // return true  if (x == mat[mid][0])  return true;    // if x is greater than first element of mid row  // store the mid row and search in lower half  if (x > mat[mid][0]) {  row = mid;  lo = mid + 1;  }    // if x is smaller than first element of mid row  // search in upper half  else  hi = mid - 1;  }    // if x is smaller than all elements of mat[][]  if (row == -1)  return false;  return search(mat[row] x); } int main() {  vector<vector<int>> mat = {{1 5 9} {14 20 21} {30 34 43}};  int x = 14;  if (searchMatrix(mat x))  cout << 'true';  else  cout << 'false';  return 0; }
Java 
class GfG {    // function to binary search for x in arr[]  static boolean search(int[] arr int x) {  int lo = 0 hi = arr.length - 1;  while (lo <= hi) {  int mid = (lo + hi) / 2;  if (x == arr[mid])  return true;  if (x < arr[mid])  hi = mid - 1;  else  lo = mid + 1;  }  return false;  }    // function to search element x in fully   // sorted matrix  static boolean searchMatrix(int[][] mat int x) {  int n = mat.length m = mat[0].length;  int lo = 0 hi = n - 1;  int row = -1;  while (lo <= hi) {  int mid = (lo + hi) / 2;  // if the first element of mid row is equal to x  // return true  if (x == mat[mid][0])  return true;  // if x is greater than first element of mid row  // store the mid row and search in lower half  if (x > mat[mid][0]) {  row = mid;  lo = mid + 1;  }  // if x is smaller than first element of mid row  // search in upper half  else  hi = mid - 1;  }  // if x is smaller than all elements of mat[][]  if (row == -1)  return false;  return search(mat[row] x);  }  public static void main(String[] args) {  int[][] mat = {  {1 5 9}  {14 20 21}  {30 34 43}  };  int x = 14;  if (searchMatrix(mat x))  System.out.println('true');  else  System.out.println('false');  } }
Python 
# function to binary search for x in arr[] def search(arr x): lo = 0 hi = len(arr) - 1 while lo <= hi: mid = (lo + hi) // 2 if x == arr[mid]: return True if x < arr[mid]: hi = mid - 1 else: lo = mid + 1 return False # function to search element x in fully  # sorted matrix def searchMatrix(mat x): n = len(mat) m = len(mat[0]) lo = 0 hi = n - 1 row = -1 while lo <= hi: mid = (lo + hi) // 2 # if the first element of mid row is equal to x # return true if x == mat[mid][0]: return True # if x is greater than first element of mid row # store the mid row and search in lower half if x > mat[mid][0]: row = mid lo = mid + 1 # if x is smaller than first element of mid row # search in upper half else: hi = mid - 1 # if x is smaller than all elements of mat[][] if row == -1: return False return search(mat[row] x) if __name__ == '__main__': mat = [[1 5 9] [14 20 21] [30 34 43]] x = 14 if searchMatrix(mat x): print('true') else: print('false')
C# 
using System; class GfG {    // function to binary search for x in arr[]  static bool Search(int[] arr int x) {  int lo = 0 hi = arr.Length - 1;  while (lo <= hi) {  int mid = (lo + hi) / 2;  if (x == arr[mid])  return true;  if (x < arr[mid])  hi = mid - 1;  else  lo = mid + 1;  }  return false;  }    // function to search element x in fully   // sorted matrix  static bool SearchMatrix(int[][] mat int x) {  int n = mat.Length m = mat[0].Length;  int lo = 0 hi = n - 1;  int row = -1;  while (lo <= hi) {  int mid = (lo + hi) / 2;  // if the first element of mid row is equal to x  // return true  if (x == mat[mid][0])  return true;  // if x is greater than first element of mid row  // store the mid row and search in lower half  if (x > mat[mid][0]) {  row = mid;  lo = mid + 1;  }  // if x is smaller than first element of mid row  // search in upper half  else  hi = mid - 1;  }  // if x is smaller than all elements of mat[][]  if (row == -1)  return false;  return Search(mat[row] x);  }  static void Main(string[] args) {  int[][] mat = new int[][] {  new int[] {1 5 9}  new int[] {14 20 21}  new int[] {30 34 43}  };  int x = 14;  if (SearchMatrix(mat x))  Console.WriteLine('true');  else  Console.WriteLine('false');  } }
JavaScript 
// function to binary search for x in arr[] function search(arr x) {  let lo = 0 hi = arr.length - 1;  while (lo <= hi) {  let mid = Math.floor((lo + hi) / 2);  if (x === arr[mid])  return true;  if (x < arr[mid])  hi = mid - 1;  else  lo = mid + 1;  }  return false; } // function to search element x in fully  // sorted matrix function searchMatrix(mat x) {  let n = mat.length m = mat[0].length;  let lo = 0 hi = n - 1;  let row = -1;  while (lo <= hi) {  let mid = Math.floor((lo + hi) / 2);  // if the first element of mid row is equal to x  // return true  if (x === mat[mid][0])  return true;  // if x is greater than first element of mid row  // store the mid row and search in lower half  if (x > mat[mid][0]) {  row = mid;  lo = mid + 1;  }  // if x is smaller than first element of mid row  // search in upper half  else  hi = mid - 1;  }  // if x is smaller than all elements of mat[][]  if (row === -1)  return false;  return search(mat[row] x); } // Driver code const mat = [  [1 5 9]  [14 20 21]  [30 34 43] ]; const x = 14; if (searchMatrix(mat x))  console.log('true'); else  console.log('false');

Uitvoer 
true

[Verwachte aanpak] Eenmalig binair zoeken gebruiken - O(log(n × m)) en O(1) Spatie

Het idee is om de gegeven matrix als 1D-array te beschouwen en slechts één binaire zoekopdracht toe te passen.
Voor een matrix met de grootte n x m, en we kunnen deze beschouwen als een 1D-array met de grootte n*m, zou de eerste index 0 zijn en de laatste index n*m-1. We moeten dus binair zoeken van laag = 0 tot hoog = (n*m-1).

Hoe vind je het element in de 2D-matrix dat overeenkomt met index = mid?

Omdat elke rij mat[][] m elementen zal hebben, kunnen we de rij van het element als (midden / m) en de kolom van het element als (midden % m) . Vervolgens kunnen we x vergelijken met arr[mid/m][mid%m] voor elk mid en onze binaire zoekopdracht voltooien.

C++ 
#include    #include  using namespace std; bool searchMatrix(vector<vector<int>>& mat int x) {  int n = mat.size() m = mat[0].size();  int lo = 0 hi = n * m - 1;  while (lo <= hi) {  int mid = (lo + hi) / 2;    // find row and column of element at mid index  int row = mid / m;  int col = mid % m;    // if x is found return true  if (mat[row][col] == x)   return true;    // if x is greater than mat[row][col] search   // in right half  if (mat[row][col] < x)   lo = mid + 1;    // if x is less than mat[row][col] search   // in left half  else   hi = mid - 1;  }  return false; } int main() {  vector<vector<int>> mat = {{1 5 9}   {14 20 21}   {30 34 43}};  int x = 14;  if (searchMatrix(mat x))  cout << 'true';  else  cout << 'false';  return 0; }
Java 
class GfG {  static boolean searchMatrix(int[][] mat int x) {  int n = mat.length m = mat[0].length;  int lo = 0 hi = n * m - 1;  while (lo <= hi) {  int mid = (lo + hi) / 2;  // find row and column of element at mid index  int row = mid / m;  int col = mid % m;  // if x is found return true  if (mat[row][col] == x)  return true;  // if x is greater than mat[row][col] search   // in right half  if (mat[row][col] < x)  lo = mid + 1;  // if x is less than mat[row][col] search   // in left half  else  hi = mid - 1;  }  return false;  }  public static void main(String[] args) {  int[][] mat = {{1 5 9}   {14 20 21}   {30 34 43}};  int x = 14;  if (searchMatrix(mat x))  System.out.println('true');  else  System.out.println('false');  } }
Python 
def searchMatrix(mat x): n = len(mat) m = len(mat[0]) lo hi = 0 n * m - 1 while lo <= hi: mid = (lo + hi) // 2 # find row and column of element at mid index row = mid // m col = mid % m # if x is found return true if mat[row][col] == x: return True # if x is greater than mat[row][col] search  # in right half if mat[row][col] < x: lo = mid + 1 # if x is less than mat[row][col] search  # in left half else: hi = mid - 1 return False if __name__ == '__main__': mat = [[1 5 9] [14 20 21] [30 34 43]] x = 14 if searchMatrix(mat x): print('true') else: print('false')
C# 
using System; class GfG {    // function to search for x in the matrix   // using binary search  static bool searchMatrix(int[] mat int x) {  int n = mat.GetLength(0) m = mat.GetLength(1);  int lo = 0 hi = n * m - 1;  while (lo <= hi) {  int mid = (lo + hi) / 2;  // find row and column of element at mid index  int row = mid / m;  int col = mid % m;  // if x is found return true  if (mat[row col] == x)  return true;  // if x is greater than mat[row col] search  // in right half  if (mat[row col] < x)  lo = mid + 1;  // if x is less than mat[row col] search   // in left half  else  hi = mid - 1;  }  return false;  }  static void Main() {  int[] mat = { { 1 5 9 } { 14 20 21 } { 30 34 43 } };  int x = 14;  if (searchMatrix(mat x))  Console.WriteLine('true');  else  Console.WriteLine('false');  } }
JavaScript 
function searchMatrix(mat x) {  let n = mat.length m = mat[0].length;  let lo = 0 hi = n * m - 1;  while (lo <= hi) {  let mid = Math.floor((lo + hi) / 2);  // find row and column of element at mid index  let row = Math.floor(mid / m);  let col = mid % m;  // if x is found return true  if (mat[row][col] === x)  return true;  // if x is greater than mat[row][col] search   // in right half  if (mat[row][col] < x)  lo = mid + 1;  // if x is less than mat[row][col] search   // in left half  else  hi = mid - 1;  }  return false; } // Driver Code let mat = [[1 5 9] [14 20 21] [30 34 43]]; let x = 14; if (searchMatrix(mat x))  console.log('true'); else  console.log('false');

Uitvoer 
true
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