A nested loop puzzle

Which of the following two code segments is faster? Assume that compiler makes no optimizations. 

C++

/* FIRST */
for(i = 0; i < 10; i++)
  for(j = 0; j < 100; j++)
    //do something

// This code is contributed by Shubham Singh

C

/* FIRST */
for(i=0;i<10;i++)
  for(j=0;j<100;j++)
    //do something

Java

/* First */

for(i = 0; i < 10; i++)
  for(j = 0; j < 100; j++)
    //do something

// This code is contributed by sarajadhav12052009

Python3

# FIRST
for i in range(10):
    for j in range(100):
        #do something

# This code is contributed by shivani

C#

// FIRST
for(i = 0; i < 10; i++)
  for(j = 0; j < 100; j++)
    //do something

// This code is contributed by aditya942003patil

Javascript

// Javascript equivalent
// Using nested for loop to loop through 10 and 100
for (let i = 0; i < 10; i++) {
    for (let j = 0; j < 100; j++) {
        // do something
    }
}

C++

/* SECOND */
for(i=0;i<100;i++)
  for(j=0;j<10;j++)
    //do something

//This code is contributed by Shubham Singh

C

/* SECOND */
for(i=0;i<100;i++)
  for(j=0;j<10;j++)
    //do something

Java

/* Second */

for(i = 0; i < 100; i++)
  for(j = 0; j < 10; j++)
    //do something

// This code is contributed by sarajadhav12052009

Python3

# SECOND
for i in range(100):
    for j in range(10):

      # Do something

# This code is contributed by shivani

C#

// SECOND
for(i=0;i<100;i++)
   for(j=0;j<10;j++)

     //do something

//This code is contributed by aditya942003patil

Javascript

/* Second */

for(let i = 0; i < 100; i++) { // use let keyword to declare i and j
for(let j = 0; j < 10; j++) {
// do something
}
}

Both code segments provide same functionality, and the code inside the two for loops would be executed same number of times in both code segments. 
If we take a closer look then we can see that the SECOND does more operations than the FIRST. It executes all three parts (assignment, comparison and increment) of the for loop more times than the corresponding parts of FIRST: 

1. The SECOND executes assignment operations ( j = 0 or i = 0) 101 times while FIRST executes only 11 times.
2. The SECOND does 101 + 1100 comparisons (i < 100 or j < 10) while the FIRST does 11 + 1010 comparisons (i < 10 or j < 100).
3. The SECOND executes 1100 increment operations (i++ or j++) while the FIRST executes 1010 increment operation.

Below code counts the number of increment operations executed in FIRST and SECOND, and prints the counts.

C++

// C++ program to count number of increment
// operations in FIRST and SECOND
#include<iostream>

using namespace std;

int main()
{
  int c1 = 0, c2 = 0;

  /* FIRST */
  for(int i=0;i<10;i++,c1++)
    for(int j=0;j<100;j++, c1++);


  /* SECOND */
  for(int i=0; i<100; i++, c2++)
      for(int j=0; j<10; j++, c2++);


  cout << " Count in FIRST = " <<c1 << endl;
  cout << " Count in SECOND  = " <<c2 << endl;

  getchar();
  return 0;
}

Java

// Java program to count number of increment
// operations in FIRST and SECOND

import java.io.*;

class GFG {
    public static void main (String[] args) {
       int c1 = 0, c2 = 0;

        for(int i=0; i<10;i++, c1++){
          for(int j=0;j<100;j++, c1++){}
        }


        for(int i=0;i<100;i++, c2++){
              for(int j=0;j<10;j++, c2++){}
        }

        System.out.println( "Count in First = "+c1);
        System.out.println( "Count in SECOND = "+c2);
    }
}

Python3

# Python program to count number of increment
# operations in FIRST and SECOND
c1 = 0
c2 = 0

# FIRST
for i in range (10):
    for j in range (100):
        c1 += 1
    c1 += 1

# SECOND
for i in range (100):
    for j in range (10):
        c2 += 1
    c2 += 1

print("Count in FIRST = " ,c1)
print("Count in SECOND  = " ,c2)

# This code is contributed by shivanisinghss2110

C#

// C# program to count number of increment
// operations in FIRST and SECOND
using System;

class GFG{

public static void Main (String[] args)
{
   int c1 = 0, c2 = 0;

    for(int i = 0; i < 10; i++, c1++)
    {
        for(int j = 0; j < 100;j++, c1++){}
    }

    for(int i = 0; i < 100; i++, c2++)
    {
        for(int j = 0; j < 10; j++, c2++){}
    }

    Console.WriteLine("Count in First = " + c1);
    Console.WriteLine("Count in SECOND = " + c2);
}
}

// This code is contributed by shivanisinghss2110

Javascript

<script>

// JavaScript program to count number of increment
// operations in FIRST and SECOND
       let c1 = 0, c2 = 0;

        for(let i=0; i<10;i++, c1++){
          for(let j=0;j<100;j++, c1++){}
        }


        for(let i=0;i<100;i++, c2++){
              for(let j=0;j<10;j++, c2++){}
        }

        document.write( "Count in First = "+c1 +"<br>");
        document.write( "Count in SECOND = "+c2);
  // this code is contributed by shivanisinghss2110

</script>

Output

 Count in FIRST = 1010
 Count in SECOND  = 1100

Below code counts the number of comparison operations executed by FIRST and SECOND

C++

//program to count the number of comparison
//operations executed by FIRST and SECOND */
#include<iostream>

using namespace std;

int main()
{
   int c1 = 0, c2 = 0;

   /* FIRST */
   for(int i=0; ++c1&&i<10; i++)
      for(int j=0; ++c1&&j<100;j++);


   /* SECOND */
   for(int i=0; ++c2&&i<100; i++)
      for(int j=0; ++c2&&j<10; j++);


   cout << " Count for FIRST  " <<c1 << endl;
   cout << " Count for SECOND  " <<c2 << endl;
   getchar();
   return 0;
}

Java

/*package whatever //do not write package name here */

import java.io.*;

class GFG {
    public static void main (String[] args) {
         int c1 = 0, c2 = 0;

        for(int i=0; i<++c1 && i<10;i++) {
              for(int j=0;j<++c1 &&j<100;j++) {
              }
        }

        for(int i=0;i<++c2 && i<100;i++) {
              for(int j=0;j<++c2 &&j<10;j++) {
              }
        }

        System.out.println( "Count in FIRST = "+c1);
        System.out.println( "Count in SECOND  = "+c2);

    }
}

Python3

#program to count the number of comparison
#operations executed by FIRST and SECOND */

# FIRST
c1 = 1
c2 = 1
i = 0
while (i < c1 and i < 10):
    j = -1
    c1 += 1
    while (j < c1 and j < 100):
        c1 += 1
        j += 1
    i += 1

# SECOND
i = 0
while (i < c2 and i < 100):
    j = -1
    c2 += 1
    while (j < c2 and j < 10):
        c2 += 1
        j += 1
    i += 1

print(" Count for FIRST  " , c1)
print(" Count for SECOND  " , c2)

# This code is contributed by shivanisinghss2110

C#

/*package whatever //do not write package name here */

using System;

class GFG {
    public static void Main (String[] args) {
         int c1 = 0, c2 = 0;

        for(int i = 0; i < ++c1 && i < 10; i++)
        {
              for(int j = 0; j < ++c1 && j < 100; j++)
              {
              }
        }

        for(int i = 0; i < ++c2 && i < 100; i++) {
              for(int j = 0; j < ++c2 && j < 10; j++) {
              }
        }

        Console.WriteLine( "Count in FIRST = "+c1);
        Console.WriteLine( "Count in SECOND  = "+c2);

    }
}

// This code is contributed by shivanisinghss2110

Javascript

<script>

/*package whatever //do not write package name here */
//program to count the number of comparison
//operations executed by FIRST and SECOND */

         let c1 = 0, c2 = 0;

        for(let i=0; i<++c1 && i<10;i++) {
              for(let j=0;j<++c1 &&j<100;j++) {
              }
        }

        for(let i=0;i<++c2 && i<100;i++) {
              for(let j=0;j<++c2 &&j<10;j++) {
              }
        }

        document.write( "Count in FIRST = "+c1 +"<br>");
        document.write( "Count in SECOND  = "+c2);

// this code is contributed by shivanisinghss2110

</script>

Output

 Count for FIRST  1021
 Count for SECOND  1201

Thanks to Dheeraj for suggesting the solution.
Please write comments if you find any of the answers/codes incorrect, or you want to share more information about the topics discussed above.