Google – Partition Array of 0 and 1


Google – Partition Array of 0 and 1
给一个数组,只包含0和1,找出一个partition的位置,使得左边0的个数和右边1的个数加起来和最大。
[Solution]
Brute Force, DP
[Solution #1]
Brute Force, 每个位置算左右两边0和1的个数.
O(n^2) time
[Solution #2]
DP,两个数组记录每个位置左边0的个数和右边1的个数。
O(3n) time, O(2n) space
[Solution #3]
DP with space optimization.
space可以优化到O(1), 用两个变量存左边0的个数和右边1的个数。leftZero从0开始,rightOne从(number of ones in the whole array)开始。从左向右扫,碰到0就leftZero++,碰到1就rightOne–.
// DP without space optimization
class Solution {
  public int partitionArray(int[] nums) {
    if (nums == null || nums.length == 0) {
      return -1;
    }
     
    int n = nums.length;
    int[] leftZeroes = new int[n];
    int[] rightOnes = new int[n];
    leftZeroes[0] = nums[0] == 0? 1 : 0;
    rightOnes[n - 1] = nums[n - 1] == 1? 1 : 0;
    for (int i = 1; i < n; i++) {
      if (nums[i] == 0) {
        leftZeroes[i] = leftZeroes[i - 1] + 1;
      }
      else {
        leftZeroes[i] = leftZeroes[i - 1];
      }
    }
     
    for (int i = n - 2; i >= 0; i--) {
      if (nums[i] == 1) {
        rightOnes[i] = rightOnes[i + 1] + 1;
      }
      else {
        rightOnes[i] = rightOnes[i + 1];
      }
    }   
     
    int result = 0;
    int maxSum = 0;
    for (int i = 0; i < n; i++) {
      if (leftZeroes[i] + rightOnes[i] > maxSum) {
        maxSum = leftZeroes[i] + rightOnes[i];
        result = i;
      }
    }
     
    return result;
  }
}
 
// O(1) space
class Solution2 {
  public int partitionArray(int[] nums) {
    if (nums == null || nums.length == 0) {
      return -1;
    }
     
    int n = nums.length;
    int leftZero = 0;
    int rightOne = 0;
    for (int i = n - 1; i >= 0; i--) {
      if (nums[i] == 1) {
        rightOne++;
      }
    }
     
    int result = -1;
    int maxSum = 0;
    for (int i= 0; i < n; i++) {
      if (leftZero + rightOne > maxSum) {
        maxSum = leftZero + rightOne;
        result = i;
      }
       
      if (nums[i] == 0) {
        leftZero++;
      }
      else {
        rightOne--;
      }
    }
     
    return result;
  }
}
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