Matrix Spiral Copy(Java Solution)

A Question on Pramp platform

There are two ways of solving this question. One is iterative way, the other is recursive way. When I doing mocking interview on Pramp, the first solution that I come up with is to solve it in iterative way like the code below.

static int[] spiralCopy(int[][] inputMatrix) {
    // your code goes here
    int m = inputMatrix.length;
    int n = inputMatrix[0].length;
    int[] ans = new int[m * n];
    int index = 0;
    
    int leftBound = 0, rightBound = n - 1;
    int upperBound = 0, lowerBound = m - 1;
    while(index < m * n){
      if(upperBound <= lowerBound){
        for(int i = leftBound; i <= rightBound; i++){
          ans[index++] = inputMatrix[upperBound][i];
        }
        upperBound++;
      }
      if(leftBound <= rightBound){
        for(int i = upperBound; i <= lowerBound; i++){
          ans[index++] = inputMatrix[i][rightBound];
        }
        rightBound--;
      }
      if(upperBound <= lowerBound){
        for(int i = rightBound; i >= leftBound; i--){
          ans[index++] = inputMatrix[lowerBound][i];
        }
        lowerBound--;
      }
      if(leftBound <= rightBound){
        for(int i = lowerBound; i >= upperBound; i--){
          ans[index++] = inputMatrix[i][leftBound];
        }
        leftBound++;
      }
    }
    return ans;
}

After mocking interview, I tried to solve it with recursive function. Then, here comes the code below.

  static int index = 0;
  static int[] spiralCopy(int[][] inputMatrix) {
    // your code goes here
    int m = inputMatrix.length;
    int n = inputMatrix[0].length;
    int[] ans = new int[m * n];
    index = 0;
    spiralOut(inputMatrix, ans, 0, m - 1, 0, n - 1);
    return ans;
  }
  static void spiralOut(int[][] inputMatrix, int[] ans, int upperbound, int lowerbound, int leftbound, int rightbound){
    if(upperbound > lowerbound || leftbound > rightbound) return;
    for(int i = leftbound; i <= rightbound; i++){
      ans[index++] = inputMatrix[upperbound][i];
    }
    for(int i = upperbound + 1; i <= lowerbound; i++){
      ans[index++] = inputMatrix[i][rightbound];
    }
    if(upperbound == lowerbound || leftbound == rightbound) return;
    for(int i = rightbound - 1; i >= leftbound; i--){
      ans[index++] = inputMatrix[lowerbound][i];
    }
    for(int i = lowerbound - 1; i >= upperbound + 1; i--){
      ans[index++] = inputMatrix[i][leftbound];
    }
    spiralOut(inputMatrix, ans, upperbound + 1, lowerbound - 1, leftbound + 1, rightbound - 1);
  }

Both solutions’ time complexity is O(m * n) and the space complexity is O(m * n).