[Solved]: A partition algorithm

Since you find it interesting to solve, I do not want to completely spoil it too much for you. So here is a big hint to solve it in linear time. You start with two indices $i$ and $j$ which are the first and last indices of central segment. Initially you set $i=0$ and $j=n$. And you compute the 3 sums in $s_1$, $s_2$ and $s_3$. So initially $s_1=s_3=0$. Then you progressively increase $i$ and decrease $j$, by variations of 1, keeping $s_1$ and $s_3$ about the same value (by doing the change on the smallest sum $s_1$ or $s_3$), until one becomes bigger than $s_2$. Each increment or decrement takes constant time (adding or substracting the value of an element to two of the sums, and doing a few comparisons). When one of $s_1$ and $s_3$ becomes larger than $s_2$, say for example $s_1$. Then you have $s_1geq s_2 > s_3$. Then you know that the right value for $i$ is either the current value or the previous value. Typically it is the previous value of i, if $s_1$ has become greater that the previous value of $s_2$. But things are a bit more subtle. So you note this value of $s_1$ as $m_1$, a possibility for the maximum, and then try to see if you could have achieved a lower value for the maximum with the previous value of $i$ and $s_1$, just before the last increase. So you revert $i$ to that previous value, and you start to decrease $j$ to get the lowest possible maximum for $s_2$ and $s_3$ without touching $i$ (that is easy, and recall that $s_1$ remains lower than both others sums). Let $m_{2,3}$ be the lowest maximum found. Now, if $m_{2,3}$ is greater than $m_1$, then the answer is $m_1$, else the answer is $m_{2,3}$. And you must consider the symmetrical case when $s_3$ is the first to exceed $s_2$.