Find element using minimum segments in Seven Segment Display


A Seven Segment Display can be used to to display numbers. Given an array of n natural numbers. The task is to find the number in the array which is using minimum segments to display number. If multiple numbers have a minimum number of segments, output the number having the smallest index.

Seven Segment Display

Examples :

Input : arr[] = { 1, 2, 3, 4, 5 }.
Output : 1
1 uses on 2 segments to display.

Input : arr[] = { 489, 206, 745, 123, 756 }.
Output : 745

Precompute the number of segment used by digits from 0 to 9 and store it. Now for each element of the array sum the number of segment used by each digit. Then find the element which is using the minimum number of segments.

The number of segment used by digit:
0 -> 6
1 -> 2
2 -> 5
3 -> 5
4 -> 4
5 -> 5
6 -> 6
7 -> 3
8 -> 7
9 -> 6

Below is the implementation of this approach:

C++

// C++ program to find minimum number of segments
// required
#include<bits/stdc++.h>
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(int arr[], int n)
{
    // Initalising 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];
}
  
// Driven Program
int main()
{
    int arr[] = {489, 206, 745, 123, 756};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << elementMinSegment(arr, n) << endl;
    return 0;
}

Java

// Java program to find minimum
// number of segments required
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)
{
    // Initalising 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];
}
  
    // Driver program 
    static public void main (String[] args)
    {
       int []arr = {489, 206, 745, 123, 756};
       int n = arr.length;
       System.out.println(elementMinSegment(arr, n));
    }
}
  
//This code is contributed by vt_m.

C#

// C# program to find minimum 
// number of segments required
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)
{
    // Initalising 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];
}
  
    // Driver program
    static public void Main()
    {
       int []arr = {489, 206, 745, 123, 756};
       int n = arr.Length;
       Console.WriteLine(elementMinSegment(arr, n));
    }
}
  
//This code is contributed by vt_m.

PHP

<?php
// PHP program to find minimum 
// number of segments required
  
// Precomputed values of segment
// used by digit 0 to 9.
  
$seg = array(6, 2, 5, 5, 4, 
             5, 6, 3, 7, 6);
  
// Return the number of
// segments used by x.
function computeSegment($x)
{
    global $seg;
    if ($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];
        $x = (int)$x / 10;
    }
  
    return $count;
}
  
function elementMinSegment($arr, $n)
{
    // Initalising 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 = 1; $i < $n; $i++)
    {
        $temp = computeSegment($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 = array (489, 206, 745, 123, 756);
$n = sizeof($arr);
echo elementMinSegment($arr, $n) ,"\n";
  
// This code is contributed by ajit
?>


Output :

745

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : vt_m, jit_t




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