Minimum Segments in Seven Segment Display
Last Updated :
26 Mar, 2025
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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 that uses the minimum number of segments to display the 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
Explanation: The element that uses the minimum number of segments is 1 ( i.e. 2 segments)Input : arr[] = { 489, 206, 745, 123, 756 }.
Output : 745
Explanation: The element with smallest index that uses the minimum number of segments is 745 ( i.e. 12 segments)
The idea is to 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
#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(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;
}
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));
}
}
# 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));
}
}
// 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));
Output
745
Time Complexity: O(n * log10n)
Auxiliary Space: O(10)
