Two Furthest Houses With Different Colors

Two Furthest Houses With Different Colors

There are n houses evenly lined up on the street, and each house is beautifully painted. You are given a 0-indexed integer array colors of length n, where colors[i] represents the color of the ith house.

Return the maximum distance between two houses with different colors.

The distance between the ith and jth houses is abs(i - j), where abs(x) is the absolute value of x.

Example 1:
Input: colors = [1,1,1,6,1,1,1]
Output: 3
Explanation: In the above image, color 1 is blue, and color 6 is red.
The furthest two houses with different colors are house 0 and house 3.
House 0 has color 1, and house 3 has color 6. The distance between them is abs(0 - 3) = 3.
Note that houses 3 and 6 can also produce the optimal answer.

Example 2:
Input: colors = [1,8,3,8,3]
Output: 4
Explanation: In the above image, color 1 is blue, color 8 is yellow, and color 3 is green.
The furthest two houses with different colors are house 0 and house 4.
House 0 has color 1, and house 4 has color 3. The distance between them is abs(0 - 4) = 4.

Example 3:
Input: colors = [0,1]
Output: 1
Explanation: The furthest two houses with different colors are house 0 and house 1.
House 0 has color 0, and house 1 has color 1. The distance between them is abs(0 - 1) = 1.

/**
 * @param {number[]} colors
 * @return {number}
 */
var maxDistance = function(colors) {
  let start = 0;
  let end = 1;
  let maxFromStart = 0;
  let maxFromEnd = 0;
  while (end < colors.length) {
    if (colors[start] !== colors[end]) {
      let diff = Math.abs(start - end);
      if (maxFromStart < diff) {
        maxFromStart = diff;
      }
    }
    end++;
  }

  end = colors.length - 1;
  while (start <= end) {
    if (colors[start] !== colors[end]) {
      let diff = Math.abs(start - end);
      if (maxFromEnd < diff) {
        maxFromEnd = diff;
      }
    }
    start++;
  }
return Math.max(maxFromEnd, maxFromStart)
};

console.log(maxDistance([4, 4, 4, 11, 4, 4, 11, 4, 4, 4, 4, 4]));