Awesome Strings Question - Compressed Bracket Sequence

I need your help in understand the line paragraph of the editorial of

The next observation is that if we fix the number of opening brackets taken from 𝑐𝑙cl, then we can count the number of brackets that we should take from π‘π‘Ÿcr. Using these observations we can calculate the answer for π‘™β€¦π‘Ÿl…r using the following formula: π‘šπ‘–π‘›(𝑐𝑙,π‘π‘Ÿβˆ’π‘π‘Žπ‘™π‘Žπ‘›π‘π‘’)βˆ’π‘šπ‘Žπ‘₯(1,π‘šπ‘–π‘›π΅π‘Žπ‘™π‘Žπ‘›π‘π‘’)+1min(cl,crβˆ’balance)βˆ’max(1,minBalance)+1.

Could someone elaborate this formula possibly provide an example to understand it better?
AC Solution:

In the solution, please explain the left and right equations i.e; what exactly they are calculating

static long solve(int []c, int n){
    long ans=0;
    for(int l=0;l<n;l+=2){
      long min_balance=c[l];
      long balance=c[l];
      for(int r=l+1;r<n;r++){
          // adding the open brackets to cur
          long left=Math.max(0, balance-c[r]);
          long right=Math.min(min_balance, Math.min(balance-1, c[l]-1));

          // removing closed brackets from open
          min_balance=Math.min(min_balance, balance);
    return ans;