XBee - Need efficient way to regulate input voltage - 3.3v Power supply

I am struggling to find a highly efficient power source for my XBee Pro 900 Digimesh. They have incredible low sleep drains but supplying them with power going through a regulator still kills the battery even with a “really” efficient one like a Dimension Engineering 3.3v - uses 15ma at no load. So would this work - The XBee is powered by 2 alkaline batteries during sleep, when it wakes up pin DI-9 (sleep indicator) powers a transistor or relay that cuts power from the alkalines and turns on a larger lithium battery which goes through a 3.3v regulator to power the XBee while its awake. When it goes to sleep the alkalines are put back online. I assume I would need a capacitor to provide power during the switch over too. Is there an easier way and i’m just overthinking this or is this a reasonable solution to use the least power possible? I haven’t tried using just a single cell LiPo connected directly to the XBee, I wasn’t brave enough to test the XBee max V input. Thank you for any help or links you can provide.

I am using a MCP1700 regulator that uses almost no power when the Xbee is in sleep mode. Very low loss of voltage drop .05 to .1. I would think then you could just run off of the Lipo. I am not an expert but that is what I am doing.

I am also working with the Xbee Pro 900 and to for the voltage regulation I am using TPS781 series voltage regulator because it consumes very low current when at no load state. The current consumption at no load is 1uA that is very less and moreover the drop out voltage is 200mV at 150mA. So it is suitable for application so you should try.

electronic pcb assembly

I use the pro-HP900 with two 18650 batteries in series (7.4 V/4200ma) taken from the damaged laptops. I mounted the regulator circuit with the MCP1702 - (http://ww1.microchip.com/downloads/en/DeviceDoc/22008E.pdf), 2.0 µA Quiescent Current/250 mA output current. Was the best option I’ve found to save the batteries without programming.