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class Solution {
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  public int minimumDeletions(int[] nums) {
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    int max = Integer.MIN_VALUE;
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    int min = Integer.MAX_VALUE;
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    int minInd = 0;
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    int maxInd = 0;
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    int n = nums.length;
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    // First Find out the max and min element index
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    for (int i = 0; i < n; i++) {
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      if (nums[i] > max) {
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        max = nums[i];
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        maxInd = i;
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      }
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      if (nums[i] < min) {
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        min = nums[i];
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        minInd = i;
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      }
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    }
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    // if both index are same then return the part in which less number of elements are there
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    if (maxInd == minInd) {
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      return Math.min(maxInd + 1, n - maxInd);
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    }
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    // max element is right side of min element
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    if (maxInd > minInd) {
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      int count =
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          Math.min(
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              maxInd + 1,
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              n - minInd); // min of all the elements till max element and all the elements to the
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      // right of min element
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      int count1 = minInd + 1 + (n - maxInd); // all elements to the left of min and right of max
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      return Math.min(count, count1); // min of both
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    }
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    // min element is right side of the max element
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    else {
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      int count = Math.min(minInd + 1, n - maxInd);
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      int count1 = maxInd + 1 + (n - minInd);
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      return Math.min(count, count1);
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    }
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  }
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}