Dado un arreglo arr[] de N elementos, la tarea es encontrar la longitud del subarreglo más pequeño que tiene la secuencia {0, 1, 2, 3, 4} como subsecuencia.
Ejemplos:
Entrada: arr[] = {0, 1, 2, 3, 4, 2, 0, 3, 4}
Salida: 5
El subarreglo requerido es {0, 1, 2, 3, 4} con longitud mínima.
La array completa también incluye la secuencia
, pero no tiene una longitud mínima.Entrada: arr[] = {0, 1, 1, 0, 1, 2, 0, 3, 4}
Salida: 6
Acercarse:
- Mantenga una array pref[] de tamaño 5 (igual al tamaño de la secuencia) donde pref[i] almacena el recuento de i en la array dada hasta ahora.
- Podemos aumentar la cuenta de preferencia para cualquier número solo si pref[Array[i] – 1] > 0. Esto se debe a que, para tener la secuencia completa como una subsecuencia de la array, todos los elementos anteriores de la array La secuencia debe ocurrir antes que la actual. Además, almacena los índices de estos elementos encontrados hasta el momento.
- Cada vez que presenciamos 4 , es decir, el posible final de la subsecuencia y pref[3] > 0 implica que hemos encontrado la secuencia en nuestra array. Ahora marque ese índice como el final así como el punto de inicio y para todos los demás números en secuencia del 3 al 0 . Aplique la búsqueda binaria para encontrar el índice más cercano al siguiente elemento de la secuencia, lo que nos dará el tamaño del subarreglo válido actual.
- La respuesta es el tamaño mínimo de todos los subconjuntos válidos encontrados en el paso anterior.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; #define MAX_INT 1000000 // Function to return the minimum length // of a sub-array which contains // {0, 1, 2, 3, 4} as a sub-sequence int solve(int Array[], int N) { // To store the indices where 0, 1, 2, // 3 and 4 are present vector<int> pos[5]; // To store if there exist a valid prefix // of sequence in array int pref[5] = { 0 }; // Base Case if (Array[0] == 0) { pref[0] = 1; pos[0].push_back(0); } int ans = MAX_INT; for (int i = 1; i < N; i++) { // If current element is 0 if (Array[i] == 0) { // Update the count of 0s till now pref[0]++; // Push the index of the new 0 pos[0].push_back(i); } else { // To check if previous element of the // given sequence is found till now if (pref[Array[i] - 1] > 0) { pref[Array[i]]++; pos[Array[i]].push_back(i); // If it is the end of sequence if (Array[i] == 4) { int end = i; int start = i; // Iterate for other elements of the sequence for (int j = 3; j >= 0; j--) { int s = 0; int e = pos[j].size() - 1; int temp = -1; // Binary Search to find closest occurrence // less than equal to starting point while (s <= e) { int m = (s + e) / 2; if (pos[j][m] <= start) { temp = pos[j][m]; s = m + 1; } else { e = m - 1; } } // Update the starting point start = temp; } ans = min(ans, end - start + 1); } } } } return ans; } // Driver code int main() { int Array[] = { 0, 1, 2, 3, 4, 2, 0, 3, 4 }; int N = sizeof(Array) / sizeof(Array[0]); cout << solve(Array, N); return 0; }
Java
// Java implementation of the approach class GFG { static int MAX_INT = 1000000; // Function to return the minimum length // of a sub-array which contains // {0, 1, 2, 3, 4} as a sub-sequence static int solve(int[] array, int N) { // To store the indices where 0, 1, 2, // 3 and 4 are present int[][] pos = new int[5][10000]; // To store if there exist a valid prefix // of sequence in array int[] pref = new int[5]; // Base Case if (array[0] == 0) { pref[0] = 1; pos[0][pos[0].length - 1] = 0; } int ans = MAX_INT; for (int i = 1; i < N; i++) { // If current element is 0 if (array[i] == 0) { // Update the count of 0s till now pref[0]++; // Push the index of the new 0 pos[0][pos[0].length - 1] = i; } else { // To check if previous element of the // given sequence is found till now if (pref[array[i] - 1] > 0) { pref[array[i]]++; pos[array[i]][pos[array[i]].length - 1] = i; // If it is the end of sequence if (array[i] == 4) { int end = i; int start = i; // Iterate for other elements of the sequence for (int j = 3; j >= 0; j--) { int s = 0; int e = pos[j].length - 1; int temp = -1; // Binary Search to find closest occurrence // less than equal to starting point while (s <= e) { int m = (s + e) / 2; if (pos[j][m] <= start) { temp = pos[j][m]; s = m + 1; } else e = m - 1; } // Update the starting point start = temp; } ans = Math.min(ans, end - start + 1); } } } } return ans; } // Driver Code public static void main(String[] args) { int[] array = { 0, 1, 2, 3, 4, 2, 0, 3, 4 }; int N = array.length; System.out.println(solve(array, N)); } } // This code is contributed by // sanjeev2552
Python3
# Python3 implementation of the approach MAX_INT=1000000 # Function to return the minimum length # of a sub-array which contains # 0, 1, 2, 3, 4 as a sub-sequence def solve(Array, N): # To store the indices where 0, 1, 2, # 3 and 4 are present pos=[[] for i in range(5)] # To store if there exist a valid prefix # of sequence in array pref=[0 for i in range(5)] # Base Case if (Array[0] == 0): pref[0] = 1 pos[0].append(0) ans = MAX_INT for i in range(N): # If current element is 0 if (Array[i] == 0): # Update the count of 0s till now pref[0]+=1 # Push the index of the new 0 pos[0].append(i) else : # To check if previous element of the # given sequence is found till now if (pref[Array[i] - 1] > 0): pref[Array[i]]+=1 pos[Array[i]].append(i) # If it is the end of sequence if (Array[i] == 4) : end = i start = i # Iterate for other elements of the sequence for j in range(3,-1,-1): s = 0 e = len(pos[j]) - 1 temp = -1 # Binary Search to find closest occurrence # less than equal to starting point while (s <= e): m = (s + e) // 2 if (pos[j][m] <= start) : temp = pos[j][m] s = m + 1 else : e = m - 1 # Update the starting point start = temp ans = min(ans, end - start + 1) return ans # Driver code Array = [ 0, 1, 2, 3, 4, 2, 0, 3, 4] N = len(Array) print(solve(Array, N)) # This code is contributed by mohit kumar 29
C#
// C# implementation of the approach using System; class GFG { static int MAX_INT = 1000000; // Function to return the minimum length // of a sub-array which contains // {0, 1, 2, 3, 4} as a sub-sequence static int solve(int[] array, int N) { // To store the indices where 0, 1, 2, // 3 and 4 are present int[,] pos = new int[5,10000]; // To store if there exist a valid prefix // of sequence in array int[] pref = new int[5]; // Base Case if (array[0] == 0) { pref[0] = 1; pos[0,pos.GetLength(0)- 1] = 0; } int ans = MAX_INT; for (int i = 1; i < N; i++) { // If current element is 0 if (array[i] == 0) { // Update the count of 0s till now pref[0]++; // Push the index of the new 0 pos[0,pos.GetLength(0) - 1] = i; } else { // To check if previous element of the // given sequence is found till now if (pref[array[i] - 1] > 0) { pref[array[i]]++; pos[array[i],pos.GetLength(1) - 1] = i; // If it is the end of sequence if (array[i] == 4) { int end = i; int start = i; // Iterate for other elements of the sequence for (int j = 3; j >= 0; j--) { int s = 0; int e = pos.GetLength(1) - 1; int temp = -1; // Binary Search to find closest occurrence // less than equal to starting point while (s <= e) { int m = (s + e) / 2; if (pos[j,m] <= start) { temp = pos[j,m]; s = m + 1; } else e = m - 1; } // Update the starting point start = temp; } ans = Math.Min(ans, end - start + 1); } } } } return ans; } // Driver Code public static void Main(String[] args) { int[] array = { 0, 1, 2, 3, 4, 2, 0, 3, 4 }; int N = array.Length; Console.WriteLine(solve(array, N)); } } // This code is contributed by PrinciRaj1992
Javascript
<script> // Javascript implementation of the approach let MAX_INT = 1000000; // Function to return the minimum length // of a sub-array which contains // {0, 1, 2, 3, 4} as a sub-sequence function solve(array,N) { // To store the indices where 0, 1, 2, // 3 and 4 are present let pos = new Array(5); for(let i=0;i<5;i++) { pos[i]=new Array(10000); for(let j=0;j<10000;j++) { pos[i][j]=0; } } // To store if there exist a valid prefix // of sequence in array let pref = new Array(5); for(let i=0;i<5;i++) { pref[i]=0; } // Base Case if (array[0] == 0) { pref[0] = 1; pos[0][pos[0].length - 1] = 0; } let ans = MAX_INT; for (let i = 1; i < N; i++) { // If current element is 0 if (array[i] == 0) { // Update the count of 0s till now pref[0]++; // Push the index of the new 0 pos[0][pos[0].length - 1] = i; } else { // To check if previous element of the // given sequence is found till now if (pref[array[i] - 1] > 0) { pref[array[i]]++; pos[array[i]][pos[array[i]].length - 1] = i; // If it is the end of sequence if (array[i] == 4) { let end = i; let start = i; // Iterate for other elements of the sequence for (let j = 3; j >= 0; j--) { let s = 0; let e = pos[j].length - 1; let temp = -1; // Binary Search to find closest occurrence // less than equal to starting point while (s <= e) { let m = Math.floor((s + e) / 2); if (pos[j][m] <= start) { temp = pos[j][m]; s = m + 1; } else e = m - 1; } // Update the starting point start = temp; } ans = Math.min(ans, end - start + 1); } } } } return ans; } // Driver Code let array=[0, 1, 2, 3, 4, 2, 0, 3, 4]; let N = array.length; document.write(solve(array, N)); // This code is contributed by patel2127 </script>
Producción:
5