Gegeven een binaire array met grootte n waar n > 3 . Een waarde waar (of 1) in de array betekent actief en onwaar (of 0) betekent inactief. Gegeven een getal k is het de taak om het aantal actieve en inactieve cellen na k dagen te vinden. Na elke dag wordt de status van de cel actief als de linker- en rechtercel niet hetzelfde zijn, en inactief als de linker- en rechtercel hetzelfde zijn (beide 0 of beide 1).
Omdat er geen cellen vóór de meest linkse en na de meest rechtse cellen staan, worden de waardecellen vóór de meest linkse en na de meest rechtse cellen altijd als 0 (of inactief) beschouwd.
Voorbeelden:
Input : cells[] = {1 0 1 1} k = 2 Output : Active cells = 3 Inactive cells = 1 After 1 day cells[] = {0 0 1 1} After 2 days cells[] = {0 1 1 1} Input : cells[] = {0 1 0 1 0 1 0 1} k = 3 Output: Active Cells = 2 Inactive Cells = 6 Explanation : After 1 day cells[] = {1 0 0 0 0 0 0 0} After 2 days cells[] = {0 1 0 0 0 0 0 0} After 3 days cells[] = {1 0 1 0 0 0 0 0} Input : cells[] = {0 1 1 1 0 1 1 0} k = 4 Output: Active Cells = 3 Inactive Cells = 5 Het enige belangrijke is om ervoor te zorgen dat we een kopie van een bepaalde array behouden, omdat we eerdere waarden nodig hebben om de volgende dag bij te werken. Hieronder vindt u gedetailleerde stappen.
- Eerst kopiëren we de cellen[]-array naar temp[]-array en brengen we wijzigingen aan in temp[]-array volgens de gegeven voorwaarde.
- In de voorwaarde wordt gegeven dat als de linker- en rechtercel van de cel de volgende dag inactief of actief zijn, ik inactief wordt, d.w.z.; (cellen[i-1] == 0 en cellen[i+1] == 0) of (cellen[i-1] == 1 en cellen[i+1] == 1) dan cellen[i] = 0. Deze voorwaarden kunnen worden toegepast met behulp van XOR van cellen[i-1] en cellen[i+1].
- Voor 0'de indexceltemperatuur[0] = 0^cellen[1] en voor (n-1)'de indexceltemperatuur[n-1] = 0^cellen[n-2].
- Voer nu voor index 1 tot n-2 de volgende bewerking uit: temp[i] = cellen[i-1] ^ cellen[i+1]
- Herhaal het proces totdat k dagen zijn voltooid.
Hieronder volgt de implementatie van bovenstaande stappen.
C++
// C++ program to count active and inactive cells after k // days #include
// Java program to count active and // inactive cells after k days class GFG { // cells[] - store current status // of cells n - Number of cells // temp[] - to perform intermediate operations // k - number of days // active - count of active cells after k days // inactive - count of active cells after k days static void activeAndInactive(boolean cells[] int n int k) { // copy cells[] array into temp [] array boolean temp[] = new boolean[n]; for (int i = 0; i < n; i++) temp[i] = cells[i]; // Iterate for k days while (k-- > 0) { // Finding next values for corner cells temp[0] = false ^ cells[1]; temp[n - 1] = false ^ cells[n - 2]; // Compute values of intermediate cells // If both cells active or inactive then // temp[i]=0 else temp[i] = 1. for (int i = 1; i <= n - 2; i++) temp[i] = cells[i - 1] ^ cells[i + 1]; // Copy temp[] to cells[] for next iteration for (int i = 0; i < n; i++) cells[i] = temp[i]; } // count active and inactive cells int active = 0 inactive = 0; for (int i = 0; i < n; i++) if (cells[i] == true) active++; else inactive++; System.out.print('Active Cells = ' + active + ' ' + 'Inactive Cells = ' + inactive); } // Driver code public static void main(String[] args) { boolean cells[] = {false true false true false true false true}; int k = 3; int n = cells.length; activeAndInactive(cells n k); } } // This code is contributed by Anant Agarwal.
# Python program to count # active and inactive cells after k # days # cells[] - store current # status of cells # n - Number of cells # temp[] - to perform # intermediate operations # k - number of days # active - count of active # cells after k days # inactive - count of active # cells after k days def activeAndInactive(cellsnk): # copy cells[] array into temp [] array temp=[] for i in range(n+1): temp.append(False) for i in range(n): temp[i] = cells[i] # Iterate for k days while (k >0): # Finding next values for corner cells temp[0] = False^cells[1] temp[n-1] = False^cells[n-2] # Compute values of intermediate cells # If both cells active or # inactive then temp[i]=0 # else temp[i] = 1. for i in range(1n-2+1): temp[i] = cells[i-1] ^ cells[i+1] # Copy temp[] to cells[] # for next iteration for i in range(n): cells[i] = temp[i] k-=1 # count active and inactive cells active = 0 inactive = 0; for i in range(n): if(cells[i] == True): active+=1 else: inactive+=1 print('Active Cells ='active' ' 'Inactive Cells =' inactive) # Driver code cells = [False True False True False True False True] k = 3 n =len(cells) activeAndInactive(cells n k) # This code is contributed # by Anant Agarwal.
// C# program to count active and // inactive cells after k days using System; class GFG { // cells[] - store current status // of cells n - Number of cells // temp[] - to perform intermediate // operations k - number of days // active - count of active cells // after k days inactive - count // of active cells after k days static void activeAndInactive(bool []cells int n int k) { // copy cells[] array into // temp [] array bool []temp = new bool[n]; for (int i = 0; i < n; i++) temp[i] = cells[i]; // Iterate for k days while (k-- > 0) { // Finding next values // for corner cells temp[0] = false ^ cells[1]; temp[n - 1] = false ^ cells[n - 2]; // Compute values of intermediate cells // If both cells active or inactive then // temp[i]=0 else temp[i] = 1. for (int i = 1; i <= n - 2; i++) temp[i] = cells[i - 1] ^ cells[i + 1]; // Copy temp[] to cells[] // for next iteration for (int i = 0; i < n; i++) cells[i] = temp[i]; } // count active and inactive cells int active = 0 inactive = 0; for (int i = 0; i < n; i++) if (cells[i] == true) active++; else inactive++; Console.Write('Active Cells = ' + active + ' ' + 'Inactive Cells = ' + inactive); } // Driver code public static void Main() { bool []cells = {false true false true false true false true}; int k = 3; int n = cells.Length; activeAndInactive(cells n k); } } // This code is contributed by Nitin Mittal.
// PHP program to count active // and inactive cells after k // days // cells[] - store current status // of cells n - Number of cells // temp[] - to perform intermediate // operations k - number of days // active - count of active cells // after k days inactive - count of // active cells after k days function activeAndInactive($cells $n $k) { // copy cells[] array into // temp [] array $temp = array(); for ($i = 0; $i < $n ; $i++) $temp[$i] = $cells[$i]; // Iterate for k days while ($k--) { // Finding next values // for corner cells $temp[0] = 0 ^ $cells[1]; $temp[$n - 1] = 0 ^ $cells[$n - 2]; // Compute values of // intermediate cells // If both cells active // or inactive then temp[i]=0 // else temp[i] = 1. for ($i = 1; $i <= $n - 2; $i++) $temp[$i] = $cells[$i - 1] ^ $cells[$i + 1]; // Copy temp[] to cells[] // for next iteration for ($i = 0; $i < $n; $i++) $cells[$i] = $temp[$i]; } // count active and // inactive cells $active = 0;$inactive = 0; for ($i = 0; $i < $n; $i++) ($cells[$i] == 1)? $active++ : $inactive++; echo 'Active Cells = ' $active ' Inactive Cells = ' $inactive; } // Driver Code $cells= array(0 1 0 1 0 1 0 1); $k = 3; $n = count($cells); activeAndInactive($cells $n $k); // This code is contributed by anuj_67. ?> <script> // javascript program to count active and // inactive cells after k days // cells - store current status // of cells n - Number of cells // temp - to perform intermediate operations // k - number of days // active - count of active cells after k days // inactive - count of active cells after k days function activeAndInactive(cells n k) { // copy cells array into temp array var temp = Array(n).fill(false); for (i = 0; i < n; i++) temp[i] = cells[i]; // Iterate for k days while (k-- > 0) { // Finding next values for corner cells temp[0] = false ^ cells[1]; temp[n - 1] = false ^ cells[n - 2]; // Compute values of intermediate cells // If both cells active or inactive then // temp[i]=0 else temp[i] = 1. for (i = 1; i <= n - 2; i++) temp[i] = cells[i - 1] ^ cells[i + 1]; // Copy temp to cells for next iteration for (i = 0; i < n; i++) cells[i] = temp[i]; } // count active and inactive cells var active = 0 inactive = 0; for (i = 0; i < n; i++) if (cells[i] == true) active++; else inactive++; document.write('Active Cells = ' + active + ' ' + 'Inactive Cells = ' + inactive); } // Driver code var cells = [ false true false true false true false true ]; var k = 3; var n = cells.length; activeAndInactive(cells n k); // This code is contributed by Rajput-Ji </script>
Uitvoer
Active Cells = 2 Inactive Cells = 6
Tijdcomplexiteit: O(N*K) waarbij N de grootte van een array is en K het aantal dagen.
Hulpruimte: O(N)
Dit artikel is beoordeeld door team geeksforgeeks. Als u een betere aanpak voor dit probleem heeft, deel deze dan.