Generate All Permutations of Array in DSA C - Execution Trace

Concept Flow - Generate All Permutations of Array

Start with full array

↓

Choose element at index i

↓

Swap element i with element j (j >= i)

↓

Recurse with i+1

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If i == array length - 1

↓

Print current permutation

↓

Backtrack: Swap back elements i and j

↩Back to swap step for next j

We pick each element in turn, swap it with the current position, recurse to fix the next position, print when done, then swap back to restore.

Execution Sample

DSA C

void permute(int *arr, int start, int end) {
  if (start == end) {
    print(arr, end + 1);
  } else {
    for (int i = start; i <= end; i++) {
      swap(arr[start], arr[i]);
      permute(arr, start+1, end);
      swap(arr[start], arr[i]);
    }
  }
}

This code generates all permutations by swapping elements starting from 'start' index and recursing.

Execution Table

Step	Operation	Indices Swapped	Current Array State	Action
1	Start permute	-	[1, 2, 3]	Call permute(arr, 0, 2)
2	Swap	(0,0)	[1, 2, 3]	Swap elements at 0 and 0
3	Recurse	-	[1, 2, 3]	Call permute(arr, 1, 2)
4	Swap	(1,1)	[1, 2, 3]	Swap elements at 1 and 1
5	Recurse	-	[1, 2, 3]	Call permute(arr, 2, 2)
6	Print	-	[1, 2, 3]	Print permutation
7	Backtrack	(1,1)	[1, 2, 3]	Swap back elements at 1 and 1
8	Swap	(1,2)	[1, 3, 2]	Swap elements at 1 and 2
9	Recurse	-	[1, 3, 2]	Call permute(arr, 2, 2)
10	Print	-	[1, 3, 2]	Print permutation
11	Backtrack	(1,2)	[1, 2, 3]	Swap back elements at 1 and 2
12	Backtrack	(0,0)	[1, 2, 3]	Swap back elements at 0 and 0
13	Swap	(0,1)	[2, 1, 3]	Swap elements at 0 and 1
14	Recurse	-	[2, 1, 3]	Call permute(arr, 1, 2)
15	Swap	(1,1)	[2, 1, 3]	Swap elements at 1 and 1
16	Recurse	-	[2, 1, 3]	Call permute(arr, 2, 2)
17	Print	-	[2, 1, 3]	Print permutation
18	Backtrack	(1,1)	[2, 1, 3]	Swap back elements at 1 and 1
19	Swap	(1,2)	[2, 3, 1]	Swap elements at 1 and 2
20	Recurse	-	[2, 3, 1]	Call permute(arr, 2, 2)
21	Print	-	[2, 3, 1]	Print permutation
22	Backtrack	(1,2)	[2, 1, 3]	Swap back elements at 1 and 2
23	Backtrack	(0,1)	[1, 2, 3]	Swap back elements at 0 and 1
24	Swap	(0,2)	[3, 2, 1]	Swap elements at 0 and 2
25	Recurse	-	[3, 2, 1]	Call permute(arr, 1, 2)
26	Swap	(1,1)	[3, 2, 1]	Swap elements at 1 and 1
27	Recurse	-	[3, 2, 1]	Call permute(arr, 2, 2)
28	Print	-	[3, 2, 1]	Print permutation
29	Backtrack	(1,1)	[3, 2, 1]	Swap back elements at 1 and 1
30	Swap	(1,2)	[3, 1, 2]	Swap elements at 1 and 2
31	Recurse	-	[3, 1, 2]	Call permute(arr, 2, 2)
32	Print	-	[3, 1, 2]	Print permutation
33	Backtrack	(1,2)	[3, 2, 1]	Swap back elements at 1 and 2
34	Backtrack	(0,2)	[1, 2, 3]	Swap back elements at 0 and 2
35	End	-	[1, 2, 3]	All permutations generated

💡 All swaps and recursions completed, original array restored

Variable Tracker

Variable	Start	After Step 2	After Step 8	After Step 13	After Step 19	After Step 24	After Step 30	Final
arr	[1,2,3]	[1,2,3]	[1,3,2]	[2,1,3]	[2,3,1]	[3,2,1]	[3,1,2]	[1,2,3]
start	0	1	1	1	1	1	1	-
end	2	2	2	2	2	2	2	-

Key Moments - 3 Insights

Why do we swap back the elements after recursion?

Swapping back restores the array to its original state before the next iteration, as shown in steps like 7, 11, 22, 23, 33, and 34 in the execution_table.

Why do we recurse only when start < end?

When start == end (step 5, 9, 16, 20, 27, 31), we have fixed all positions and print the permutation. No further recursion is needed.

How does swapping elements generate all permutations?

By swapping each element into the current 'start' position and recursing, we explore all possible orders. The execution_table shows different swaps at steps 2, 8, 13, 19, 24, and 30.

Visual Quiz - 3 Questions

Look at the execution_table at step 8, what is the array state?

A [1, 2, 3]

B [2, 1, 3]

C [1, 3, 2]

D [3, 2, 1]

💡 Hint

Check the 'Current Array State' column at step 8 in execution_table

At which step does the first permutation get printed?

A Step 5

B Step 6

C Step 7

D Step 8

💡 Hint

Look for 'Print' operation in execution_table

If we skip swapping back after recursion, what happens to the array state?

A Array stays changed, permutations will be incorrect

B Array resets automatically, no effect

C Only first permutation prints correctly

D Program crashes

💡 Hint

Refer to key_moments about why swapping back is needed and steps like 7 and 11

Concept Snapshot

Generate All Permutations of Array:
- Use recursion with a 'start' index
- Swap elements at 'start' with each index >= start
- Recurse with start+1
- Print when start == end
- Swap back after recursion to restore array
- This explores all possible orders of elements

Full Transcript

This concept shows how to generate all permutations of an array by swapping elements recursively. We start from the first index, swap it with itself and others, then recurse to fix the next index. When we reach the last index, we print the current array as a permutation. After each recursion, we swap back to restore the array for the next iteration. The execution table traces each swap, recursion, print, and backtrack step, showing how the array changes. Key moments clarify why swapping back is necessary and how recursion stops when the start index reaches the end. The visual quiz tests understanding of array states and recursion steps. This method ensures all permutations are generated without duplicates or missing any order.