NAJMANJŠE ŠTEVILO SEGMENTOV V SEDEMSEGMENTNEM PRIKAZU

Za prikaz številk lahko uporabite sedemsegmentni zaslon. Glede na niz n naravna števila. Naloga je najti število v matriki, ki uporablja najmanjše število segmentov za prikaz števila. Če ima več števil najmanjše število segmentov, izpišite število z najmanjšim indeksom.

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Primeri:

Vnos: arr[] = { 1 2 3 4 5 }.
Izhod: 1
Pojasnilo: Element, ki uporablja najmanjše število segmentov, je 1 (tj. 2 segmenta)
Vnos: arr[] = { 489 206 745 123 756 }.
Izhod: 745
Pojasnilo: Element z najmanjšim indeksom, ki uporablja najmanjše število segmentov, je 745 (tj. 12 segmentov)

Ideja je vnaprej izračunati število segmentov, ki jih uporabljajo števke od 0 do 9, in ga shraniti. Zdaj za vsak element matrike seštejte število segmentov, ki jih uporablja posamezna števka. Nato poiščite element, ki uporablja najmanjše število segmentov.

Številka segmenta, ki ga uporablja cifra:
0 -> 6
1 -> 2
2 -> 5
3 -> 5
4 -> 4
5 -> 5
6 -> 6
7 -> 3
8 -> 7
9 -> 6

C++

#include   using namespace std; // Precomputed values of segment used by digit 0 to 9. const int seg[10] = { 6 2 5 5 4 5 6 3 7 6}; // Return the number of segments used by x. int computeSegment(int x) {  if (x == 0)  return seg[0];  int count = 0;  // Finding sum of the segment used by  // each digit of a number.  while (x)  {  count += seg[x%10];  x /= 10;  }  return count; } int elementMinSegment(vector<int> arr int n) {  // Initialising the minimum segment and minimum  // number index.  int minseg = computeSegment(arr[0]);  int minindex = 0;  // Finding and comparing segment used  // by each number arr[i].  for (int i = 1; i < n; i++)  {  int temp = computeSegment(arr[i]);  // If arr[i] used less segment then update  // minimum segment and minimum number.  if (temp < minseg)  {  minseg = temp;  minindex = i;  }  }  return arr[minindex]; } int main() {  vector<int> arr = {489 206 745 123 756};  int n = arr.size();   cout << elementMinSegment(arr n) << endl;  return 0; }

Java

import java.io.*; class GFG {   // Precomputed values of segment  // used by digit 0 to 9. static int []seg = { 6 2 5 5 4 5 6 3 7 6}; // Return the number of segments used by x. static int computeSegment(int x) {  if (x == 0)  return seg[0];  int count = 0;  // Finding sum of the segment used by  // each digit of a number.  while (x > 0)  {  count += seg[x % 10];  x /= 10;  }  return count; } static int elementMinSegment(int []arr int n) {  // Initialising the minimum segment   // and minimum number index.  int minseg = computeSegment(arr[0]);  int minindex = 0;  // Finding and comparing segment used  // by each number arr[i].  for (int i = 1; i < n; i++)  {  int temp = computeSegment(arr[i]);  // If arr[i] used less segment then update  // minimum segment and minimum number.  if (temp < minseg)  {  minseg = temp;  minindex = i;  }  }  return arr[minindex]; }  static public void main (String[] args)  {  int []arr = {489 206 745 123 756};  int n = arr.length;  System.out.println(elementMinSegment(arr n));  } }

Python

# Precomputed values of segment # used by digit 0 to 9. seg = [6 2 5 5 4 5 6 3 7 6] # Return the number of # segments used by x. def computeSegment(x): if(x == 0): return seg[0] count = 0 # Finding sum of the segment  # used by each digit of a number. while(x): count += seg[x % 10] x = x // 10 return count # function to return minimum sum index def elementMinSegment(arr n): # Initialising the minimum  # segment and minimum number index. minseg = computeSegment(arr[0]) minindex = 0 # Finding and comparing segment # used by each number arr[i]. for i in range(1 n): temp = computeSegment(arr[i]) # If arr[i] used less segment # then update minimum segment # and minimum number. if(temp < minseg): minseg = temp minindex = i return arr[minindex] # Driver Code arr = [489 206 745 123 756] n = len(arr) # function print required answer print(elementMinSegment(arr n)) # This code is contributed by # Sanjit_Prasad

using System; class GFG{   // Precomputed values of segment // used by digit 0 to 9. static int []seg = new int[10]{ 6 2 5 5 4  5 6 3 7 6}; // Return the number of segments used by x. static int computeSegment(int x) {  if (x == 0)  return seg[0];  int count = 0;  // Finding sum of the segment used by  // each digit of a number.  while (x > 0)  {  count += seg[x % 10];  x /= 10;  }  return count; } static int elementMinSegment(int []arr int n) {  // Initialising the minimum segment  // and minimum number index.  int minseg = computeSegment(arr[0]);  int minindex = 0;  // Finding and comparing segment used  // by each number arr[i].  for (int i = 1; i < n; i++)  {  int temp = computeSegment(arr[i]);  // If arr[i] used less segment then update  // minimum segment and minimum number.  if (temp < minseg)  {  minseg = temp;  minindex = i;  }  }  return arr[minindex]; }  static public void Main()  {  int []arr = {489 206 745 123 756};  int n = arr.Length;  Console.WriteLine(elementMinSegment(arr n));  } }

JavaScript

// Precomputed values of segment // used by digit 0 to 9. let seg = [ 6 2 5 5 4 5 6 3 7 6]; // Return the number of segments used by x. function computeSegment(x) {  if (x == 0)  return seg[0];  let count = 0;  // Finding sum of the segment used by  // each digit of a number.  while (x > 0)  {  count += seg[x % 10];  x = parseInt(x / 10 10);  }  return count; } function elementMinSegment(arr n) {    // Initialising the minimum segment  // and minimum number index.  let minseg = computeSegment(arr[0]);  let minindex = 0;  // Finding and comparing segment used  // by each number arr[i].  for(let i = 1; i < n; i++)  {  let temp = computeSegment(arr[i]);  // If arr[i] used less segment then update  // minimum segment and minimum number.  if (temp < minseg)  {  minseg = temp;  minindex = i;  }  }  return arr[minindex]; } // Driver code let arr = [ 489 206 745 123 756 ]; let n = arr.length; console.log(elementMinSegment(arr n));

Izhod

Časovna zapletenost: O(n * log ₁₀ n)
Pomožni prostor: O(10)

Ustvari kviz

TechCodeview

Najmanjše število segmentov v sedemsegmentnem prikazu