Find element using minimum segments in Seven Segment Display

• Difficulty Level : Basic
• Last Updated : 08 Jun, 2021

A Seven Segment Display can be used 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. 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

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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#includeusing namespace std; // Precomputed values of segment used by digit 0 to 9.const int seg = { 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;     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){    // Initialising the minimum segment and minimum    // number index.    int minseg = computeSegment(arr);    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 Programint main(){    int arr[] = {489, 206, 745, 123, 756};    int n = sizeof(arr)/sizeof(arr);    cout << elementMinSegment(arr, n) << endl;    return 0;}

Java

 // Java program to find minimum// number of segments requiredimport 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;     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);    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.

Python3

 # Python implementation of# the above approach # 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     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 indexdef elementMinSegment(arr, n):     # Initialising the minimum    # segment and minimum number index.    minseg = computeSegment(arr)    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 Codearr = [489, 206, 745, 123, 756]n = len(arr) # function print required answerprint(elementMinSegment(arr, n)) # This code is contributed by# Sanjit_Prasad

C#

 // C# program to find minimum// number of segments requiredusing System; class GFG{     // Precomputed values of segment// used by digit 0 to 9.static int []seg = new int{ 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;     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);    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.



Javascript



Output :

745

Time Complexity: O(n * log10n)

Auxiliary Space: O(10)
This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.