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package g0501_0600.s0598_range_addition_ii;
// #Easy #Array #Math #2022_08_25_Time_0_ms_(100.00%)_Space_44.3_MB_(71.22%)
public class Solution {
/*
* Since the incrementing starts from zero to op[0] and op[1], we only need to find the range that
* has the most overlaps. Thus we keep finding the minimum of both x and y.
*/
public int maxCount(int m, int n, int[][] ops) {
int x = m;
int y = n;
for (int[] op : ops) {
x = Math.min(x, op[0]);
y = Math.min(y, op[1]);
}
return x * y;
}
}
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