做个基本思路可以用 brute force, 但时间复杂度较高. 因为起始值都为0, 所以肯定是左上角的重合的最小的长方形就是结果, 所以我们求x, y 的最小值, 最后返回x*y.

Code     T: O(m*n)    S: O(1)

class Solution:
def rangeAddition2(self, m, n, ops):
for op1, op2 in ops:
m = min(m, op1)
n = min(n, op2)
return m*n

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